One of the most obvious applications of derivatives is to help us understand the shape of the graph of a function. In this section we will use our accumulated knowledge of derivatives to identify the most important qualitative features of graphs \(y=f(x)\text{.}\) The goal of this section is to highlight features of the graph \(y=f(x)\) that are easily
The domain of the function — take note of values where \(f\) does not exist. If the function is rational, look for where the denominator is zero. Similarly be careful to look for roots of negative numbers or other possible sources of discontinuities.
Vertical asymptotes — look for values of \(x\) at which \(f(x)\) blows up. If \(f(x)\) approaches either \(+\infty\) or \(-\infty\) as \(x\) approaches \(a\) (or possibly as \(x\) approaches \(a\) from one side) then \(x=a\) is a vertical asymptote to \(y=f(x)\text{.}\) When \(f(x)\) is a rational function (written so that common factors are cancelled), then \(y=f(x)\) has vertical asymptotes at the zeroes of the denominator.
Horizontal asymptotes — examine the limits of \(f(x)\) as \(x\to+\infty\) and \(x\to-\infty\text{.}\) Often \(f(x)\) will tend to \(+\infty\) or to \(-\infty\) or to a finite limit \(L\text{.}\) If, for example, \(\lim\limits_{x\rightarrow+\infty}f(x)=L\text{,}\) then \(y=L\) is a horizontal asymptote to \(y=f(x)\) as \(x\rightarrow\infty\text{.}\)
Since the function is rational and its denominator is zero at \(x=-3,+2\) it will have vertical asymptotes at \(x=-3,+2\text{.}\) To determine the shape around those asymptotes we need to examine the limits
Notice that when \(x\) is close to \(-3\text{,}\) the factors \((x+1)\) and \((x-2)\) are both negative, so the sign of \(f(x) = \frac{x+1}{x-2} \cdot \frac{1}{x+3}\) is the same as the sign of \(x+3\text{.}\) Hence
Finally since the numerator has degree 1 and the denominator has degree 2, we see that as \(x \to \pm \infty\text{,}\)\(f(x) \to 0\text{.}\) So \(y=0\) is a horizontal asymptote.
Since we know the behaviour around the asymptotes and we know the locations of the intercepts (as shown in the left graph below), we can then join up the pieces and smooth them out to get the a good sketch of this function (below right).
Subsection3.6.2First Derivative — Increasing or Decreasing
Now we move on to the first derivative, \(f'(x)\text{.}\) This is a good time to revisit the mean-value theorem (Theorem 2.13.5) and some of its consequences (Corollary 2.13.12). There we considered any function \(f(x)\) that is continuous on an interval \(A\le x\le B\) and differentiable on \(A\lt x\lt B\text{.}\) Then
if \(f'(x) \gt 0\) for all \(A \lt x \lt B\text{,}\) then \(f(x)\) is increasing on \([A,B]\)
Thus the sign of the derivative indicates to us whether the function is increasing or decreasing. Further, as we discussed in Section 3.5.1, we should also examine points at which the derivative is zero — critical points — and points where the derivative does not exist. These points may indicate a local maximum or minimum.
We will now consider a function \(f(x)\) that is defined on an interval \(I\text{,}\)except possibly at finitely many points of \(I\text{.}\) If \(f\) or its derivative \(f'\) is not defined at a point \(a\) of \(I\text{,}\) then we call \(a\) a singular point 1
This is the extension of the definition of “singular point” mentioned in the footnote in Definition 3.5.6.
Singular points — determine where \(f'(x)\) is not defined. If \(f'(x)\) approaches \(\pm\infty\) as \(x\) approaches a singular point \(a\text{,}\) then \(f\) has a vertical tangent there when \(f\) approaches a finite value as \(x\) approaches \(a\) (or possibly approaches \(a\) from one side) and a vertical asymptote when \(f(x)\) approaches \(\pm\infty\) as \(x\) approaches \(a\) (or possibly approaches \(a\) from one side).
Increasing and decreasing — where is the derivative positive and where is it negative. Notice that in order for the derivative to change sign, it must either pass through zero (a critical point) or have a singular point. Thus neighbouring regions of increase and decrease will be separated by critical and singular points.
We can also determine where the function is positive or negative since we know it is continuous everywhere and zero at \(x=0,6\text{.}\) Thus we must examine the intervals
When \(x \lt 0\text{,}\)\(x^3 \lt 0\) and \(x-6 \lt 0\) so \(f(x) = x^3(x-6) = (\text{negative})(\text{negative})
\gt 0\text{.}\) Similarly when \(x \gt 6\text{,}\)\(x^3 \gt 0, x-6 \gt 0\) we must have \(f(x) \gt 0\text{.}\) Finally when \(0 \lt x \lt 6\text{,}\)\(x^3 \gt 0\) but \(x-6 \lt 0\) so \(f(x) \lt 0\text{.}\) Thus
Since the function is a polynomial, it does not have any singular points, but it does have two critical points at \(x=0, 9/2\text{.}\) These two critical points split the real line into 3 open intervals
The second derivative \(f''(x)\) tells us the rate at which the derivative changes. Perhaps the easiest way to understand how to interpret the sign of the second derivative is to think about what it implies about the slope of the tangent line to the graph of the function. Consider the following sketches of \(y=1+x^2\) and \(y=-1-x^2\text{.}\)
In the case of \(y = f(x) = 1+x^2\) , \(f''(x) = 2 \gt 0\text{.}\) Notice that this means the slope, \(f'(x)\text{,}\) of the line tangent to the graph at \(x\) increases as \(x\) increases. Looking at the figure on the left above, we see that the graph always lies above the tangent lines.
For \(y = f(x) = -1-x^2\) , \(f''(x) = -2 \lt 0\text{.}\) The slope, \(f'(x)\text{,}\) of the line tangent to the graph at \(x\) decreases as \(x\) increases. Looking at the figure on the right above, we see that the graph always lies below the tangent lines.
Both of their derivatives, \(-\frac{1}{2}x^{-3/2}\) and \(-\frac{1}{2}(4-x)^{-1/2}\text{,}\) are negative, so they are decreasing functions. Examining second derivatives shows some differences.
For the first function, \(y''(x) = \frac{3}{4}x^{-5/2} \gt 0\text{,}\) so the slopes of tangent lines are increasing with \(x\) and the graph lies above its tangent lines.
However, the second function has \(y''(x) = -\frac{1}{4}(4-x)^{-3/2} \lt 0\) so the slopes of the tangent lines are decreasing with \(x\) and the graph lies below its tangent lines.
Let \(f(x)\) be a continuous function on the interval \([a,b]\) and suppose its first and second derivatives exist on that interval.
If \(f''(x) \gt 0\) for all \(a \lt x \lt b\text{,}\) then the graph of \(f\) lies above its tangent lines for \(a \lt x \lt b\) and it is said to be concave up.
If \(f''(x) \lt 0\) for all \(a \lt x \lt b\text{,}\) then the graph of \(f\) lies below its tangent lines for \(a \lt x \lt b\) and it is said to be concave down.
If \(f''(c)=0\) for some \(a \lt c \lt b\text{,}\) and the concavity of \(f\) changes across \(x=c\text{,}\) then we call \((c,f(c))\) an inflection point.
Thus the second derivative is zero (and potentially changes sign) at \(x=0,3\text{.}\) Thus we should consider the sign of the second derivative on the following intervals
Example3.6.5.Optional — \(y=x^{1/3}\) and \(y = x^{2/3}\).
In our Definition 3.6.3, concerning concavity and inflection points, we considered only functions having first and second derivatives on the entire interval of interest. In this example, we will consider the functions
We shall see that \(x=0\) is a singular point for both of those functions. There is no universal agreement as to precisely when a singular point should also be called an inflection point. We choose to extend our definition of inflection point in Definition 3.6.3 as follows. If
the function \(f(x)\) is defined and continuous on an interval \(a\lt x\lt b\) and if
then we say that \(\big(c\,,\,f(c)\big)\) is an inflection point of \(y=f(x)\text{.}\) Now let’s check out \(y=f(x)\) and \(y=g(x)\) from this point of view.
Features of \(y=f(x)\) and \(y=g(x)\) that are read off of \(f(x)\) and \(g(x)\text{:}\)
Since \(f(0)=0^{1/3}=0\) and \(g(0)=0^{2/3}=0\text{,}\) the origin \((0,0)\) lies on both \(y=f(x)\) and \(y=g(x)\text{.}\)
For example, \(1^3=1\) and \((-1)^3=-1\) so that the cube root of \(1\) is \(1^{1/3}=1\) and the cube root of \(-1\) is \((-1)^{1/3}=-1\text{.}\) In general,
Consequently the graph \(y=f(x)=x^{1/3}\) lies below the \(x\)-axis when \(x\lt 0\) and lies above the \(x\)-axis when \(x>0\text{.}\) On the other hand, the graph \(y=g(x)=x^{2/3}=\big[x^{1/3}\big]^2\) lies on or above the \(x\)-axis for all \(x\text{.}\)
So the graph \(y=f(x)\) is increasing on both sides of the singular point \(x=0\text{,}\) while the graph \(y=g(x)\) is decreasing to the left of \(x=0\) and is increasing to the right of \(x=0\text{.}\) As \(x\rightarrow 0\text{,}\)\(f'(x)\) and \(g'(x)\) become infinite. That is, the slopes of the tangent lines at \(\big(x,f(x)\big)\) and \(\big(x,g(x)\big)\) become infinite and the tangent lines become vertical.
So the graph \(y=g(x)\) is concave down on both sides of the singular point \(x=0\text{,}\) while the graph \(y=f(x)\) is concave up to the left of \(x=0\) and is concave down to the right of \(x=0\text{.}\)
Since the concavity changes at \(x=0\) for \(y=f(x)\text{,}\) but not for \(y=g(x)\text{,}\)\((0,0)\) is an inflection point for \(y=f(x)\text{,}\) but not for \(y=g(x)\text{.}\) We have the following sketch for \(y=f(x)=x^{1/3}\)
Note that the curve \(y=f(x)=x^{1/3}\) looks perfectly smooth, even though \(f'(x)\rightarrow\infty\) as \(x\rightarrow 0\text{.}\) There is no kink or discontinuity at \((0,0)\text{.}\) The singularity at \(x=0\) has caused the \(y\)-axis to be a vertical tangent to the curve, but has not prevented the curve from looking smooth.
Before we proceed to some examples, we should examine some simple symmetries possessed by some functions. We’ll look at three symmetries — evenness, oddness and periodicity. If a function possesses one of these symmetries then it can be exploited to reduce the amount of work required to sketch the graph of the function.
Notice that the points \((x_0,y_0)\) and \((-x_0,y_0)\) are just reflections of each other across the \(y\)-axis. Consequently, to draw the graph \(y=f(x)\text{,}\) it suffices to draw the part of the graph with \(x\ge 0\) and then reflect it in the \(y\)–axis. Here is an example. The part with \(x\ge 0\) is on the left and the full graph is on the right.
Now the symmetry is a little harder to interpret pictorially. To get from \((x_0,y_0)\) to \((-x_0,-y_0)\) one can first reflect \((x_0,y_0)\) in the \(y\)–axis to get to \((-x_0,y_0)\) and then reflect the result in the \(x\)–axis to get to \((-x_0,-y_0)\text{.}\) Consequently, to draw the graph \(y=f(x)\text{,}\) it suffices to draw the part of the graph with \(x\ge 0\) and then reflect it first in the \(y\)–axis and then in the \(x\)–axis. Here is an example. First, here is the part of the graph with \(x\ge 0\text{.}\)
The \(y\)-intercept is \(g(0) = \frac{-9}{3} = -3\text{.}\) And \(x\)-intercepts are given by the solution of \(x^2-9=0\text{,}\) namely \(x=\pm 3\text{.}\) Note that we only need to establish \(x=3\) as an intercept. Then since \(g\) is even, we know that \(x=-3\) is also an intercept.
We can already produce a quite reasonable sketch just by putting in the horizontal asymptote and the intercepts and drawing a smooth curve between them.
Note that we have drawn the function as never crossing the asymptote \(y=1\text{,}\) however we have not yet proved that. We could by trying to solve \(g(x)=1\text{.}\)
\begin{align*}
\frac{x^2-9}{x^2+3} &= 1\\
x^2-9 &= x^2+3\\
-9=3 & \text{ so no solutions.}
\end{align*}
Alternatively we could analyse the first derivative to see how the function approaches the asymptote.
There are no singular points since the denominator is nowhere zero. The only critical point is at \(x=0\text{.}\) Thus we must find the sign of \(g'(x)\) on the intervals
When \(x \gt 0\text{,}\)\(24x \gt 0\) and \((x^2+3) \gt 0\text{,}\) so \(g'(x) \gt 0\) and the function is increasing. By even symmetry we know that when \(x \lt 0\) the function must be decreasing. Hence the critical point \(x=0\) is a local minimum of the function.
Notice that since the function is increasing for \(x \gt 0\) and the function must approach the horizontal asymptote \(y=1\) from below. Thus the sketch above is quite accurate.
It is clear that \(g''(x) = 0\) when \(x=\pm 1\text{.}\) Note that, again, we can infer the zero at \(x=-1\) from the zero at \(x=1\) by the even symmetry. Thus we need to examine the sign of \(g''(x)\) the intervals
When \(|x| \lt 1\) we have \((1-x^2) \gt 0\) so that \(g''(x) \gt 0\) and the function is concave up. When \(|x| \gt 1\) we have \((1-x^2) \lt 0\) so that \(g''(x) \lt 0\) and the function is concave down. Thus the points \(x=\pm 1\) are inflection points. Their coordinates are \((\pm1, g(\pm1)) =(\pm 1,-2)\text{.}\)
More generally \(f(x+kP)=f(x)\) for all integers \(k\text{.}\) Thus if \(f\) has period \(P\text{,}\) then it also has period \(nP\) for all natural numbers \(n\text{.}\) The smallest period is called the fundamental period.
Note that the point \((x_0+P,y_0)\) can be obtained by translating \((x_0,y_0)\) horizontally by \(P\text{.}\) Similarly the point \((x_0+nP,y_0)\) can be found by repeatedly translating \((x_0,y_0)\) horizontally by \(P\text{.}\)
Consequently, to draw the graph \(y=f(x)\text{,}\) it suffices to draw one period of the graph, say the part with \(0\le x\le P\text{,}\) and then translate it repeatedly. Here is an example. Here is a sketch of one period
Above we have described how we can use our accumulated knowledge of derivatives to quickly identify the most important qualitative features of graphs \(y=f(x)\text{.}\) Here we give the reader a quick checklist of things to examine in order to produce an accurate sketch based on properties that are easily read off from \(f(x)\text{,}\)\(f'(x)\) and \(f''(x)\text{.}\)
Next compute \(f(0)\text{,}\)\(\lim_{x\rightarrow\infty} f(x)\) and \(\lim_{x\rightarrow-\infty} f(x)\) and look for solutions to \(f(x)=0\) that you can easily find. Then
The \(y\)-intercept is \(y=1\text{.}\) The \(x\)-intercepts are not easily computed since it is a cubic polynomial that does not factor nicely 2
With the aid of a computer we can find the \(x\)-intercepts numerically: \(x\approx -1.879385242, 0.3472963553\text{,}\) and \(1.532088886\text{.}\) If you are interested in more details check out Appendix C.
. So for this example we don’t worry about finding them.
The critical points (where \(f'(x)=0\)) are at \(x=\pm 1\text{.}\) Further since the derivative is a polynomial it is defined everywhere and there are no singular points. The critical points split the real line into the intervals \((-\infty,-1),(-1,1)\) and \((1,\infty)\text{.}\)
The second derivative is zero when \(x=0\text{,}\) and the problem is quite easy to analyse. Clearly, \(f''(x) \lt 0\) when \(x \lt 0\) and \(f''(x) \gt 0\) when \(x \gt 0\text{.}\)
The critical points are at \(x=0,3\text{.}\) Since the function is a polynomial there are no singular points. The critical points split the real line into the intervals \((-\infty,0)\text{,}\)\((0,3)\) and \((3,\infty)\text{.}\)
Thus the function is convex up for \(x \lt 0\text{,}\) then convex down for \(0 \lt x \lt 2\text{,}\) and finally convex up again for \(x \gt 2\text{.}\) Hence \((0,f(0))=(0,0)\) and \((2,f(2))=(2,-16)\) are inflection points.
The critical points are at \(x=1,3\text{.}\) Since the function is a polynomial there are no singular points. The critical points split the real line into the intervals \((-\infty,1)\text{,}\)\((1,3)\) and \((3,\infty)\text{.}\)
Since \(f(-x) = \dfrac{-x}{x^2-4} = - f(x)\text{,}\) it is odd. Indeed this means that we only need to examine what happens to the function for \(x \geq 0\) and we can then infer what happens for \(x\leq 0\) using \(f(-x) = -f(x)\text{.}\) In practice we will sketch the graph for \(x\geq0\) and then infer the rest from this symmetry.
The \(y\)-intercept is \(y=f(0)=0\text{,}\) while the \(x\)-intercepts are given by the solution of \(f(x)=0\text{.}\) So the only \(x\)-intercept is \(0\text{.}\)
Since \(f\) is rational, it may have vertical asymptotes where its denominator is zero — at \(x=\pm 2\text{.}\) Since the function is odd, we only have to analyse the asymptote at \(x=2\) and we can then infer what happens at \(x=-2\) by symmetry.
Hence there are no critical points. There are singular points where the denominator is zero, namely \(x=\pm2\text{.}\) Before we proceed, notice that the numerator is always negative and the denominator is always positive. Hence \(f'(x) \lt 0\) except at \(x=\pm 2\) where it is undefined.
So \(f''(x)=0\) when \(x=0\) and does not exist when \(x=\pm 2\text{.}\) This splits the real line into the intervals \((-\infty,-2), (-2,0), (0,2)\) and \((2,\infty)\text{.}\) However we only need to consider \(x \geq 0\) (because of the odd symmetry).
This final example is more substantial since the function has singular points (points where the derivative is undefined). The analysis is more involved.
The function is the cube root of a rational function. The rational function is defined except at \(x=6\text{,}\) so the domain of \(f\) is all reals except \(x=6\text{.}\)
That is, the line \(y=1\) will be a horizontal asymptote to the graph \(y=f(x)\) both for \(x\rightarrow+\infty\) and for \(x\rightarrow-\infty\text{.}\)
Our function \(f(x)\rightarrow+\infty\) as \(x\rightarrow 6\text{,}\) because of the \((1-6/x)^2\) in its denominator. So \(y=f(x)\) has \(x=6\) as a vertical asymptote.
Notice that the derivative is nowhere equal to zero, so the function has no critical points. However there are two places the derivative is undefined. The terms
are undefined at \(x=6,0\) respectively. Hence \(x=0,6\) are singular points. These split the real line into the intervals \((-\infty,0), (0,6)\) and \((6,\infty)\text{.}\)
When \(x \lt 0\text{,}\)\((x-6) \lt 0\text{,}\) we have that \((x-6)^{-\frac53} \lt 0\) and \(x^{-\frac13} \lt 0\) and so \(f'(x)=-4 \cdot
(\text{negative})\cdot(\text{negative}) \lt 0 \text{.}\)
When \(0 \lt x \lt 6\text{,}\)\((x-6) \lt 0\text{,}\) we have that \((x-6)^{-\frac53} \lt 0\) and \(x^{-\frac13} \gt 0\) and so \(f'(x) \gt 0\text{.}\)
We already know that \(x=6\) is a vertical asymptote of the function, so it is not surprising that the lines tangent to the graph become vertical as we approach 6. The behavior around \(x=0\) is less standard, since the lines tangent to the graph become vertical, but \(x=0\) is not a vertical asymptote of the function. Indeed the function takes a finite value \(y=f(0)=0\text{.}\)
Both of the factors \({\Big(\frac{1}{x-6}\Big)}^{\frac{8}{3}}
={\Big(\frac{1}{\root{3}\of{x-6}}\Big)}^8\) and \(\frac{1}{x^{\frac{4}{3}}}
=\Big(\frac{1}{\root{3}\of{x}}\Big)^4\) are even powers and so are positive (though possibly infinite). So the sign of \(f''(x)\) is the same as the sign of the factor \(x-1\text{.}\) Thus
It is hard to see the inflection point at \(x=1\text{,}\)\(y=f(1)=\frac{1}{ \root{3}\of{25} }\) in the above sketch. So here is a blow up of the part of the sketch around \(x=1\text{.}\)
On the graph below, mark the intervals where \(f''(x) \gt 0\) (i.e. \(f(x)\) is concave up) and where \(f''(x) \lt 0\) (i.e. \(f(x)\) is concave down).
Questions 3.6.7.5 through 3.6.7.7 ask you to show that certain things are true. Give a clear explanation using concepts and theorems from this textbook.
Suppose \(f(x)\) is a function whose second derivative exists and is continuous for all real numbers, and \(x=3\) is an inflection point of \(f(x)\text{.}\) Use the Intermediate Value Theorem to show that \(f''(3)=0\text{.}\)
Suppose \(f(x)\) is an even function defined for all real numbers. Below is the curve \(y=f(x)\) when \(x \gt 0\text{.}\) Complete the sketch of the curve.
Suppose \(f(x)\) is an odd function defined for all real numbers. Below is the curve \(y=f(x)\) when \(x \gt 0\text{.}\) Complete the sketch of the curve.
In Questions 3.6.7.6 and 3.6.7.7, you will sketch the graphs of functions with an exponential component. In the next section, you will learn how to find their horizontal asymptotes, but for now these are given to you.
Determine intervals where \(f(x)\) is concave upwards or downwards, and the \(x\) coordinates of inflection points (if any). You may use, without verifying it, the formula \(f''(x) = (3x -12)(3 - x)^{-3/2}/4\text{.}\)
Sketch the graph \(y = f(x)\text{,}\) showing the features given in items (a) to (d) above and giving the \((x, y)\) coordinates for all points occurring above.
Indicate the critical points, local and absolute maxima and minima, vertical and horizontal asymptotes, inflection points and regions where the curve is concave upward or downward.
Graph \(f(x)\text{.}\) Include local and absolute maxima and minima, regions where \(f(x)\) is increasing or decreasing, regions where the curve is concave upward or downward, and any asymptotes.
Sketch the graph of \(f(x)\text{.}\) Indicate the critical points, local and/or absolute maxima and minima, and asymptotes. Without actually calculating the inflection points, indicate on the graph their approximate location.
Find all inflection points and intervals of increase, decrease, convexity up, and convexity down. You may use without proof the formula \(f''(x) = (x^3-3x)e^{-x^2/2}\text{.}\)
Graph the equation \(y = f(x)\text{,}\) including all important features. (In particular, find all local maxima and minima and all inflection points.) Additionally, find the maximum and minimum values of \(f(x)\) on the interval \([0,\pi]\text{.}\)
Sketch the graph of \(y=f(x)=x^5-x\text{,}\) indicating asymptotes, local maxima and minima, inflection points, and where the graph is concave up/concave down.
Consider the function \(f(x)=x^5-x+k\text{,}\) where \(k\) is a constant, \(-\infty \lt k \lt \infty\text{.}\) How many roots does the function have? (Your answer might depend on the value of \(k\text{.}\))
Define inverse hyperbolic trigonometric functions \(\sinh^{-1}(x)\) and \(\cosh^{-1}(x)\text{,}\) carefully specifing their domains of definition. Sketch the graphs of \(\sinh^{-1}(x)\) and \(\cosh^{-1}(x)\text{.}\)
