As in the previous questions, we want to use the Squeeze Theorem. If \(x \lt 0\text{,}\) then \(-x\) is positive, so \(x \lt -x\text{.}\) Use this fact when you bound your expressions.
You’ll want to simplify this, since \(t=\frac{1}{2}\) is not in the domain of the function. One way to start your simplification is to add the fractions in the numerator by finding a common denominator.
When you’re considering \(\displaystyle\lim_{x \rightarrow -4^-} f(x)\text{,}\) you’re only considering values of \(x\) that are less than\(-4\text{.}\) When you’re considering \(\displaystyle\lim_{x \rightarrow -4^+} f(x)\text{,}\) think about the domain of the rational function in the top line.
Remember that \(\sqrt{\ }\) is defined to be the positive square root. Consequently, if \(x \lt 0\text{,}\) then \(\sqrt{x^2}\text{,}\) which is positive, is not the same as \(x\text{,}\) which is negative.
Divide both the numerator and the denominator by \(x\) (which is the largest power of \(x\) in the denominator). In the numerator, move the resulting factor of \(1/x\) inside the two roots. Be careful about the signs when you do so. Even and odd roots behave differently-- see Question 1.5.2.10.
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator. It is not always true that \(\sqrt{x^2}=x\text{.}\)
Divide both the numerator and the denominator by the highest power of \(x\) that is in the denominator. When is \(\sqrt{x}=x\text{,}\) and when is \(\sqrt{x}=-x\text{?}\)
For \(f\) to be differentiable at \(x=2\text{,}\) two things must be true: it must be continuous at \(x=2\text{,}\) and the derivative from the right must equal the derivative from the left.
From Section 1.2, compare the definition of velocity to the definition of a derivative. When you’re finding the derivative, you’ll need to cancel a lot on the numerator, which you can do by expanding the polynomials.
You’ll need to look at limits from the left and right. The fact that \(f(0)=0\) is useful for your computation. Recall that if \(x \lt 0\) then \(\sqrt{x^2}=|x|=-x\text{.}\)
A generic point on the curve has coordinates \((\alpha, \alpha^2)\text{.}\) In terms of \(\alpha\text{,}\) what is the equation of the tangent line to the curve at the point \((\alpha, \alpha^2)\text{?}\) What does it mean for \((1,-3)\) to be on that line?
First, factor an \(x\) out of the derivative. What’s left over looks like a quadratic equation, if you take \(x^2\) to be your variable, instead of \(x\text{.}\)
First simplify. Don’t be confused by the role reversal of \(x\) and \(y\text{:}\)\(x\) is just the name of the function \(\big(2y+\tfrac{1}{y}\big)\cdot y^3\text{,}\) which is a function of the variable \(y\text{.}\) You are to differentiate with respect to \(y\text{.}\)
Let \(m\) be the slope of such a tangent line, and let \(P_1\) and \(P_2\) be the points where the tangent line is tangent to the two curves, respectively. There are three equations \(m\) fulfils: it has the same slope as the curves at the given points, and it is the slope of the line passing through the points \(P_1\) and \(P_2\text{.}\)
A line has equation \(y=mx+b\text{,}\) for some constants \(m\) and \(b\text{.}\) What has to be true for \(y=mb+x\) to be tangent to the first curve at the point \(x=\alpha\text{,}\) and to the second at the point \(x=\beta\text{?}\)
In order to be differentiable, a function should be continuous. To determine the differentiability of the function at \(x=1\text{,}\) use the definition of the derivative.
The only spot to worry about is when \(x=0\text{.}\) For \(f(x)\) to be differentiable, it must be continuous, so first find the value of \(b\) that makes \(f\) continuous at \(x=0\text{.}\) Then, find the value of \(a\) that makes the derivatives from the left and right of \(x=0\) equal to each other.
You can set up the derivative using the limit definition: \(f'(0)=\ds\lim_{h \to 0}\dfrac{f(h)-f(0)}{h}\text{.}\) If the limit exists, it gives you \(f'(0)\text{;}\) if the limit does not exist, you conclude \(f'(0)\) does not exist.
Recall \(|x|=\left\{\begin{array}{rl}
x&x\ge 0\\
-x&x \lt 0
\end{array}\right.\text{.}\) To determine whether \(h(x)\) is differentiable at \(x=0\text{,}\) use the definition of the derivative.
In this chapter, we learned \(\ds\lim_{x \to 0}\dfrac{\sin x}{x}=1\text{.}\) If you divide the numerator and denominator by \(x^5\text{,}\) you can make use of this knowledge.
If \(g(x)=\cos x\) and \(h(x)=5x+3\text{,}\) then \(f(x)=g(h(x))\text{.}\) So we apply the chain rule, with “outside” function \(\cos x\) and “inside” function \(5x+3\text{.}\)
You can expand this into a polynomial, but it’s easier to use the chain rule. If \(g(x)=x^5\text{,}\) and \(h(x)=x^2+2\text{,}\) then \(f(x)=g(h(x))\text{.}\)
You can expand this into a polynomial, but it’s easier to use the chain rule. If \(g(k)=k^{17}\text{,}\) and \(h(k)=4k^4+2k^2+1\text{,}\) then \(T(k)=g(h(k))\text{.}\)
If we define \(g(x)=\sqrt{x}\) and \(h(x)=\dfrac{x^2+1}{x^2-1}\text{,}\) then \(f(x)=g(h(x))\text{.}\) To differentiate the square root function: \(\ds\diff{}{x}\{\sqrt{x}\}=\ds\diff{}{x}\left\{x^{1/2}\right\}=\dfrac{1}{2}x^{-1/2}=\dfrac{1}{2\sqrt{x}}\text{.}\)
If we let \(g(x)=\sec x\) and \({h(x)}=e^{2x+7}\text{,}\) then \(f(x)=g(h(x))\text{,}\) so by the chain rule, \(f'(x)=g'(h(x))\cdot h'(x)\text{.}\) However, in order to evaluate \(h'(x)\text{,}\) we’ll need to use the chain rule again.
The notation \(\cos^3(5x-7)\) means \(\left[\cos(5x-7)\right]^3\text{.}\) So, if \(g(x)=x^3\) and \(h(x)=\cos(5x-7)\text{,}\) then \(g(h(x))=\left[\cos(5x+7)\right]^3=\cos^3(5x+7)\text{.}\)
At time \(t\text{,}\) the particle is at the point \(\big(x(t),y(t)\big)\text{,}\) with \(x(t)=\cos t\) and \(y(t)=\sin t\text{.}\) Over time, the particle traces out a curve; let’s call that curve \(y=f(x)\text{.}\) Then \(y(t) = f\big(x(t)\big)\text{,}\) so the slope of the curve at the point \(\big(x(t),y(t)\big)\) is \(f'\big(x(t)\big)\text{.}\) You are to determine the values of \(t\) for which \(f'\big(x(t)\big)=-1\text{.}\)
This is a long, nasty problem, but it doesn’t use anything you haven’t seen before. Be methodical, and break the question into as many parts as you have to. At the end, be proud of yourself for your problem-solving abilities and tenaciousness!
Each speaker produces 3dB of noise, so if \(P\) is the power of one speaker, \(3=V(P)=10\log_{10}\left(\frac{P}{S}\right)\text{.}\) Use this to find \(V(10P)\) and \(V(100P)\text{.}\)
You’ll need to use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{.}\)
To make this easier, use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.
To make this easier, use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.
You’ll need to use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.
You’ll need to use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.
You’ll need to use logarithmic differentiation. Set \(g(x)=\log(f(x))\text{,}\) and find \(g'(x)\text{.}\) Then, use that to find \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{,}\) and again in Question 2.10.3.19.
You’ll need to use logarithmic differentiation. Differentiate \(\log(f(x))\text{,}\) then solve for \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{.}\)
You’ll need to use logarithmic differentiation. Differentiate \(\log(f(x))\text{,}\) then solve for \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{.}\)
You’ll need to use logarithmic differentiation. Differentiate \(\log(f(x))\text{,}\) then solve for \(f'(x)\text{.}\) This is the method used in the text to find \(\ds\diff{}{x} a^x\text{.}\)
Remember that \(y\) is a function of \(x\text{.}\) Use implicit differentiation, then collect all the terms containing \(\ds\diff{y}{x}\) on one side of the equation to solve for \(\ds\diff{y}{x}\text{.}\)
You can simplify the expression before you differentiate to remove the trigonometric functions. If \(\arctan x =\theta\text{,}\) then fill in the sides of the triangle below using the definition of arctangent and the Pythagorean theorem:
You can simplify the expression before you differentiate to remove the trigonometric functions. If \(\arcsin x =\theta\text{,}\) then fill in the sides of the triangle below using the definition of arctangent and the Pythagorean theorem:
To use Rolle’s Theorem, you will want two values where the function is zero. If you’re stuck finding one of them, think about when \(x^2-2\pi x\) is equal to zero.
To show that there are exactly \(n\) roots, you need to not only show that \(n\) exist, but also that there are not more than \(n\text{.}\) If you can’t explicitly find the root(s), you can use the intermediate value theorem to show they exist.
If \(f(x)=0\text{,}\) then \(|x^3|=\left|\sin\left(x^5\right)\right| \leq 1\text{.}\) When \(|x| \lt 1\text{,}\) is \(\cos(x^5)\) positive or negative?
Show that \(f\) is differentiable by showing that \(f'(x)\) exists for every \(x\text{.}\) Then, the Mean Value Theorem applies. What is the largest \(f'(x)\) can be, for any \(x\text{?}\) If \(f(100) \lt 100\text{,}\) what does the MVT tell you must be true of \(f'(c)\) for some \(c\text{?}\)
\(h'(t)\) gives the velocity of the particle, and \(h''(t)\) gives its acceleration--the rate the velocity is changing. Be wary of signs--as in legends, they may be misleading.
For 2.14.2.18.b, you know a point where the curve and tangent line intersect, and you know what the tangent line looks like. What do the derivatives tell you about the shape of the curve?
Rolle’s Theorem relates the roots of a function to the roots of its derivative. So, the fifth derivative tells us something about the fourth, the fourth derivative tells us something about the third, and so on.
You can re-write this function as a piecewise function, with branches \(x \ge 0\) and \(x \lt 0\text{.}\) To figure out the derivatives at \(x = 0\text{,}\) use the definition of a derivative.
The equation of an object falling from rest on the earth is derived in Example 3.1.2. It would be difficult to use exactly the version given for \(s(t)\text{,}\) but using the same logic, you can find an equation for the height of the flower pot at time \(t\text{.}\)
Acceleration is constant, so finding a formula for the distance your keys have travelled is a similar problem to finding a formula for something falling.
Be very careful with units. The acceleration of gravity you’re used to is \(9.8\) metres per second squared, so you might want to convert \(325\) kpm to metres per second.
First, find an equation for \(a(t)\text{,}\) the acceleration of the car, noting that \(a'(t)\) is constant. Then, use this to find an equation for the velocity of the car. Be careful about seconds versus hours.
We recommend using two different functions to describe your height: \(h_1(t)\) while you are in the air, not yet touching the trampoline, and \(h_2(t)\) while you are in the trampoline, going down.
If a trapezoid has height \(h\) and (parallel) bases \(b_1\) and \(b_2\text{,}\) then its area is \(h\left(\frac{b_1+b_2}{2}\right)\text{.}\) To figure out how wide the top of the water is when the water is at height \(h\text{,}\) you can cut the trapezoid up into a rectangle and two triangles, and make use of similar triangles.
Let \(\theta\) be the angle between the two hands. Using the Law of Cosines, you can get an expression for \(D\) in terms of \(\theta\text{.}\) To find \(\ds\diff{\theta}{t}\text{,}\) use what you know about how fast clock hands move.
The easiest way to figure out the area of the sector of an annulus (or a circle) is to figure out the area of the entire annulus, then multiply by what proportion of the entire annulus the sector is. For example, if your sector is \(\frac{1}{10}\) of the entire annulus, then its area is \(\frac{1}{10}\) of the area of the entire annulus. (See Section A.4 to see how this works out for circles.)
If you were to install the buoy, how would you choose the length of rope? For which values of \(\theta\) do \(\sin\theta\) and \(\cos\theta\) have different signs? How would those values of \(\theta\) look on the diagram?
For the question “How fast is the point moving?” in part (b), remember that the velocity of an object can be found by differentiating (with respect to time) the equation that gives the position of the object. The complicating factors in this case are that (1) the position of our object is not given as a function of time, and (2) the position of our object is given in two dimensions, not one.
(a) Since the perimeter of the cross section of the bottle does not change, \(p\) (the perimeter of the ellipse) is the same as the perimeter of the circle of radius 5.
(b) The volume of the bottle will be the area of its cross section times its height. This is always the case when you have some two-dimensional shape, and turn it into a three-dimensional object by “pulling” the shape straight up. (For example, you can think of a cylinder as a circle that has been “pulled” straight up. To understand why this formula works, think about what is means to measure the area of a shape in square centimetres, and the volume of an object in cubic centimetres.)
(c) You can use what you know about \(a\) and the formula from (a) to find \(b\) and \(\ds\diff{b}{t}\text{.}\) Then use the formula from \((b)\text{.}\)
From the text, we see the half-life of Carbon-14 is 5730 years. A microgram (\(\mu\)g) is one-millionth of a gram, but you don’t need to know that to solve this problem.
The quantity of Radium-226 in the sample at time \(t\) will be \(Q(t)=Ce^{-kt}\) for some positive constants \(C\) and \(k\text{.}\) You can use the given information to find \(C\) and \(e^{-k}\text{.}\)
The fact that the mass of the sample decreases at a rate proportional to its mass tells us that, if \(Q(t)\) is the mass of Polonium-201, the following differential equation holds:
The amount of Radium-221 in a sample at time \(t\) will be \(Q(t)=Ce^{-kt}\) for some positive constants \(C\) and \(k\text{.}\) You can leave \(C\) as a variable--it’s the original amount in the sample, which isn’t specified. What you want to find is the value of \(t\) such that \(Q(t)=0.0001Q(0)=0.0001C\text{.}\)
You don’t need to know the original amount of Polonium-210 in order to answer this question: you can leave it as some constant \(C\text{,}\) or you can call it 100%.
You can refer to Corollary 3.3.8, but you can also just differentiate the various proposed functions and see whether, in fact, \(\ds\diff{T}{t}\) is the same as \(5[T-20]\text{.}\)
where \(A\) is the ambient temperature, \(T(0)\) is the initial temperature of the copper, and \(K\) is some constant. Use the given information to find an expression for \(T(t)\) not involving any unknown constants.
The assumption that the animals grow according to the Malthusian model tells us that their population \(t\) years after 2015 is given by \(P(t)=121e^{bt}\) for some constant \(b\text{.}\)
The Malthusian model says that the population of bacteria \(t\) hours after being placed in the dish will be \(P(t)=1000e^{bt}\) for some constant \(b\text{.}\)
If the population has a net birthrate per individual per unit time of \(b\text{,}\) then the Malthusian model predicts that the number of individuals at time \(t\) will be \(P(t)=P(0)e^{bt}\text{.}\) You can use the test population to find \(e^b\text{.}\)
You’ll need some constant \(a\) to approximation \(\log(0.93) \approx \log(a)\text{.}\) This \(a\) should have two properties: it should be close to 0.93, and you should be able to easily evaluate \(\log(a)\text{.}\)
You’ll need some constant \(a\) to approximate \(\arcsin(0.1) \approx \arcsin(a)\text{.}\) This \(a\) should have two properties: it should be close to 0.1, and you should be able to easily evaluate \(\arcsin(a)\text{.}\)
You’ll need some constant \(a\) to approximate \(\sqrt{3}\tan(1) \approx \sqrt{3}\tan(a)\text{.}\) This \(a\) should have two properties: it should be close to 1, and you should be able to easily evaluate \(\sqrt{3}\tan(a)\text{.}\)
We could figure out \(10.1^3\) exactly, if we wanted, with pen and paper. Since we’re asking for an approximation, we aren’t after perfect accuracy. Rather, we’re after ease of calculation.
You’ll need to centre your approximation about some \(x=a\text{,}\) which should have two properties: you can easily compute \(\log(a)\text{,}\) and \(a\) is close to \(0.93\text{.}\)
If \(Q(x)\) is the quadratic approximation of \(f\) about \(3\text{,}\) then \(Q(3)=f(3)\text{,}\)\(Q'(3)=f'(3)\text{,}\) and \(Q''(3)=f''(3)\text{.}\)
Let \(f(x)=e^x\text{,}\) and use the quadratic approximation of \(f(x)\) about \(x=0\) (given in your text, or you can reproduce it) to approximate \(f(1)\text{.}\)
and notice that the term you want (containing \(f^{(10)}(5)\)) corresponds to \(k=10\) in the standard form, but is not the term corresponding to \(k=10\) in the polynomial given in the question.
\(\Delta x\) and \(\Delta y \) represent changes in \(x\) and \(y\text{,}\) respectively, while \(f(x)\) and \(f\left(x+\Delta x \right)\) are the \(y\)-values the function takes.
The exact area desired is \(A_0\text{.}\) Let the corresponding exact radius desired be \(r_0\text{.}\) The linear approximation tells us \(\Delta A \approx A'(r_0) \Delta r\text{.}\) Use this relationship, and what you know about the error allowable in \(A\text{,}\) to find the error allowable in \(r\text{.}\)
Remember that the amount of the isotope present at time \(t\) is \(Q(t)=Q(0)e^{-kt}\) for some constant \(k\text{.}\) The measured quantity after 3 years will allow you to replace \(k\) in the equation, then solving \(Q(t)=\frac{1}{2}Q(0)\) for \(t\) will give you the half-life of the isotope.
You are approximating a third order polynomial with a fifth order Taylor polynomial. You should be able to tell how good your approximation will be without a long calculation.
In this case, Equation 3.4.33 tells us that \(\left|f(0.1)-T_2(0.1)\right| = \left|\dfrac{f^{(3)}(c)}{3!}(0.1-0)^{3}\right|\) for some \(c\) strictly between 0 and 0.1.
In this case, Equation 3.4.33 tells us that \(\left|f(30)-T_3(30)\right| = \left|\dfrac{f^{(4)}(c)}{4!}(30-32)^{4}\right|\) for some \(c\) strictly between 30 and 32.
In our case, Equation 3.4.33 tells us \(\left|f\left(0.01\right)-T_1\left(0.01\right)\right| = \left|\dfrac{f^{(2)}(c)}{2!}\left(0.01-\frac{1}{\pi}\right)^2\right|\) for some \(c\) between 0.01 and \(\dfrac{1}{\pi}\text{.}\)
Using Equation 3.4.33, \(\left|f\left(\dfrac{1}{2}\right)-T_2\left(\dfrac{1}{2}\right)\right| = \left|\dfrac{f^{(3)}(c)}{3!}\left(\dfrac{1}{2}-0\right)^{3}\right|\) for some \(c\) in \(\left(0,\dfrac{1}{2}\right)\text{.}\)
It helps to have a formula for \(f^{(n)}(x)\text{.}\) You can figure it out by taking several derivatives and noticing the pattern, but also this has been given previously in the text.
A low order Taylor approximation will give you a good enough estimation. If you guess an order, and take that Taylor polynomial, the error will probably be less than 0.001 (but you still need to check).
For part 3.4.11.14.c, after you plug in the appropriate values to Equation 3.4.33, simplify the upper and lower bounds for \(e\) separately. In particular, for the upper bound, you’ll have to solve for \(e\text{.}\)
We’re only after local extrema, not global. Let \(f(x)\) be our function. If there is some interval around \(x=2\) where nothing is bigger than \(f(2)\text{,}\) then \(f(2)\) is a local maximum, whether or not it is a maximum overall.
One way to avoid a global minimum is to have \(\ds\lim_{x \to \infty}f(x)=-\infty\text{.}\) Since \(f(x)\) keeps getting lower and lower, there is no one value that is the lowest.
One way to decide whether a critical point \(x=c\) is a local extremum is to consider the first derivative. For example: if \(f'(x)\) is negative for all \(x\) just to the left of \(c\text{,}\) and positive for all \(x\) just to the right of \(c\text{,}\) then \(f(x)\) decreases up till \(c\text{,}\) then increases after \(c\text{,}\) so \(f(x)\) has a local minimum at \(c\text{.}\)
One way to decide whether a critical point \(x=c\) is a local extremum is to consider the first derivative. For example: if \(f'(x)\) is negative for all \(x\) just to the left of \(c\text{,}\) and positive for all \(x\) just to the right of \(c\text{,}\) then \(f(x)\) decreases up till \(c\text{,}\) then increases after \(c\text{,}\) so \(f(x)\) has a local minimum at \(c\text{.}\)
Start with a formula for travel time from \(P\) to \(B\text{.}\) You might want to assign a variable to the distance from \(A\) where your buggy first reaches the road.
Check for horizontal asymptotes by evaluating \(\ds\lim_{x \to \pm \infty}f(x)\text{,}\) and check for vertical asymptotes by finding any value of \(x\) near which \(f(x)\) blows up.
Check for horizontal asymptotes by evaluating \(\ds\lim_{x \to \pm \infty}f(x)\text{,}\) and check for vertical asymptotes by finding any value of \(x\) near which \(f(x)\) blows up.
Try allowing your graph to have horizontal asymptotes. For example, let the function get closer and closer to the \(x\)-axis (or another horizontal line) without touching it.
You must show it has at least one inflection point (try the Intermediate Value Theorem), and at most one inflection point (consider whether the second derivative is increasing or decreasing).
Since \(x=3\) is an inflection point, we know the concavity of \(f(x)\) changes at \(x=3\text{.}\) That is, there is some interval around 3, with endpoints \(a\) and \(b\text{,}\) such that
\(f''(a) \lt 0\) and \(f''(x) \lt 0\) for every \(x\) between \(a\) and 3, and
Use the IVT to show that \(f''(x)=3\) for some\(x\) between \(a\) and \(b\text{;}\) then show that this value of \(x\) can’t be anything except\(x=3\text{.}\)
You’ll find the intervals of increase and decrease. These will give you a basic outline of the behaviour of the function. Use concavity to refine your picture.
The sign of the first derivative is determined entirely by the numerator, but the sign of the second derivative depends on both the numerator and the denominator.
Since you aren’t asked to find the intervals of concavity exactly, sketch the intervals of increase and decrease, and turn them into a smooth curve. You might not get exactly the intervals of concavity that are given in the solution, but there should be the same number of intervals as the solution, and they should have the same positions relative to the local extrema.
The period of this function is \(2\pi\text{.}\) So, it’s enough to graph the curve \(y=f(x)\) over the interval \([-\pi,\pi]\text{,}\) because that figure will simply repeat.
Use trigonometric identities to write \(f''(x)=-4(4\sin^2 x + \sin x -2)\text{.}\) Then you can find where \(f''(x)=0\) by setting \(y=\sin x\) and solving \(0=4y^2+y-2\text{.}\)
Once you have the graph of a function, reflect it over the line \(y=x\) to graph its inverse. Be careful of the fact that \(f(x)\) is only defined in this problem for \(x \geq 0\text{.}\)
For (b), remember that to define an inverse of a function, we need to restrict the domain of that function to an interval where it is one-to-one. Then to graph the inverse, we can simply reflect the original function over the line \(y=x\text{.}\)
For (c), set \(y(x)=\cosh^{-1}(x)\text{,}\) so \(\cosh(y(x))=x\text{.}\) The differentiate using the chain rule. To get your final answer in terms of \(x\) (instead of \(y\)), use the identity \(\cosh^2(y)-\sinh^2(y)=1\text{.}\)
Plugging in \(x=1\) to the numerator and denominator makes both zero. This is exactly one of the indeterminate forms where l’Hôpital’s rule can be directly applied.
If it is too difficult to take a derivative for l’Hôpital’s Rule, try splitting up the function into smaller chunks and evaluating their limits independently.
Start with one application of l’Hôpital’s Rule. After that, you need to consider three distinct cases: \(k \gt 2\text{,}\)\(k \lt 2\text{,}\) and \(k=2\text{.}\)
Percentage error: \(100\left|\frac{\mbox{exact}-\mbox{approx}}{\mbox{exact}}\right|\text{.}\) Absolute error: \(|\mbox{exact}-\mbox{approx}|\text{.}\) (We’ll see these definitions again in 3.4.25.)
When \(f(x)\) is positive, its antiderivative \(F(x)\) is increasing. When \(f(x)\) is negative, its antiderivative \(F(x)\) is decreasing. When \(f(x)=0\text{,}\)\(F(x)\) has a horizontal tangent line.
For any constant \(n\neq -1\text{,}\) an antiderivative of \(x^n\) is \(\frac{1}{n+1}x^{n+1}\text{.}\) The constant \(n\) does not have to be an integer.
The derivative of \(e^{5x+11}\) is close to, but not exactly the same as, \(f(x)\text{.}\) Don’t be afraid to just make a guess. But be sure to check by differentiating your guess. If the derivative isn’t what you want, you will often still learn enough to be able to then guess the correct antiderivative.
First, find the antiderivative of \(f'(x)\text{.}\) Your answer will have an unknown \(+C\) in it. Figure out which value of \(C\) gives \(f(1)=10\text{.}\)
If \(P(t)\) is the amount of power in kW-hours the house has used since time \(t=0\text{,}\) where \(t\) is measured in hours, then the information given is that \(P'(t)=0.5\sin\left(\dfrac{\pi}{24}t\right)+0.25\) kW.
This problem is similar to Question 4.1.2.11, but you’ll have to do some harder factoring. Try getting \(f(x)\) into the form \(a\left(\dfrac{1}{\sqrt{1-(bx)^2}}\right)\) for some constants \(a\) and \(b\text{.}\)
Following Example 4.1.7 let \(V(H)\) be the volume of the solid formed by rotating the segment of the parabola from \(x=-H\) to \(x=H\text{.}\) The plan is to evaluate
and then antidifferentiate \(V'(H)\) to find \(V(H)\text{.}\) Since you don’t know \(V(H)-V(h)\) (yet), first find upper and lower bounds on it when \(h \lt H\text{.}\) These bounds can be the volumes of two cylinders, one with radius \(H\) (and what height?) and the other with radius \(h\text{.}\)
