Mathematics Program · Real Analysis II · Basic Analysis I & II · Jiří Lebl · sources and attribution

Section 4.4 Inverse function theorem

Note: less than 1 lecture (optional section, needed for Section 5.4, requires Section 3.6)
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Subsection 4.4.1 Inverse function theorem

We start with a simple example. Consider the function \(f(x) \coloneqq a x\) for a number \(a \neq 0\text{.}\) Then \(f \colon \R \to \R\) is bijective, and the inverse is \(f^{-1}(y) = \frac{1}{a} y\text{.}\) In particular, \(f'(x) = a\) and \((f^{-1})'(y) = \frac{1}{a}\text{.}\) As differentiable functions are “infinitesimally like” linear functions, we expect the same sort of behavior from the inverse of a differentiable function. The main idea of differentiating inverse functions is the following lemma.
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Proof.

By Proposition 3.6.6, \(f\) has a continuous inverse. For convenience, call the inverse \(g \colon J \to I\text{.}\) Let \(x_0,y_0\) be as in the statement. For \(x \in I\text{,}\) write \(y \coloneqq f(x)\text{.}\) If \(x \neq x_0\text{,}\) and so \(y \neq y_0\text{,}\) we find
\begin{equation} \frac{g(y)-g(y_0)}{y-y_0} = \frac{g\bigl(f(x)\bigr)-g\bigl(f(x_0)\bigr)}{f(x)-f(x_0)} = \frac{x-x_0}{f(x)-f(x_0)} . \end{equation}
See Figure 4.10 for the geometric idea.
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Two diagrams. On the left, a function f that looks like the square root is shown in the first quadrant. On the horizontal axis there is a point labeled x equals g of y. A dashed line goes vertically up from this point until it hits the corresponding point on the graph of f. Another dashed line, now horizontal goes left until it hits the y axis and a point marked f of x equals y. Another point x sub 0 equals g of y sub 0 is also marked a bit to the right of the first. The same setup with the dashed lines is repeated except instead of x it is x sub 0 and instead of y it is y sub 0. A secant line is drawn through the two points on the graph and it is labeled as having slope f of x minus f of x sub 0, the whole thing over quantity x minus x sub 0 and that equals y minus y sub 0, the whole thing over quantity g of y minus g of y sub 0. The right hand diagram is the same thing but the picture is simply reflected across the x equals y line, so the x-axis becomes the y-axis and vice-versa, and the graph now looks more like x squared. The two expressions of the tangent line are the reciprocals.
Figure 4.10. Interpretation of the derivative of the inverse function.
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Let
\begin{equation} Q(x) \coloneqq \begin{cases} \frac{x-x_0}{f(x)-f(x_0)} & \text{if } x \neq x_0, \\ \frac{1}{f'(x_0)} & \text{if } x = x_0 \quad \text{(notice that } f'(x_0) \neq 0 \text{)}. \end{cases} \end{equation}
As \(f\) is differentiable at \(x_0\text{,}\)
\begin{equation} \lim_{x \to x_0} Q(x) = \lim_{x \to x_0} \frac{x-x_0}{f(x)-f(x_0)} = \frac{1}{f'(x_0)} = Q(x_0) , \end{equation}
that is, \(Q\) is continuous at \(x_0\text{.}\) As \(g(y)\) is continuous at \(y_0\text{,}\) the composition \(Q\bigl(g(y)\bigr) = \frac{g(y)-g(y_0)}{y-y_0}\) is continuous at \(y_0\) by Proposition 3.2.7. Therefore,
\begin{equation} \frac{1}{f'\bigl(g(y_0)\bigr)} = Q\bigl(g(y_0)\bigr) = \lim_{y \to y_0} Q\bigl(g(y)\bigr) = \lim_{y \to y_0} \frac{g(y)-g(y_0)}{y-y_0} . \end{equation}
So \(g\) is differentiable at \(y_0\text{,}\) and \(g'(y_0) = \frac{1}{f'\left(\vphantom{1^1_1}g(y_0)\right)}\text{.}\)
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If \(f'\) is continuous and nonzero at all \(x \in I\text{,}\) then the lemma applies at all \(x \in I\text{.}\) As \(g\) is also continuous (it is differentiable), the derivative \(g'(y) = \frac{1}{f'\left(\vphantom{1^1_1}g(y)\right)}\) must be continuous.
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What is usually called the inverse function theorem is the following result.
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Proof.

Without loss of generality, suppose \(f'(x_0) > 0\text{.}\) As \(f'\) is continuous, there must exist an open interval \(I = (x_0-\delta,x_0+\delta)\) such that \(f'(x) > 0\) for all \(x \in I\text{.}\) See Exercise 3.2.11.
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By Proposition 4.2.8, \(f\) is strictly increasing on \(I\text{,}\) and hence the restriction \(f|_{I}\) is bijective onto \(J \coloneqq f(I)\text{.}\) As \(f\) is continuous, Corollary 3.6.3 (or directly via the intermediate value theorem) implies that \(f(I)\) is an interval. Now apply Lemma 4.4.1.
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In Example 1.2.3, we saw how difficult an endeavor was proving the existence of \(\sqrt{2}\) without any tools. With the intermediate value theorem, the existence of roots is almost trivial, and with the machinery of this section, we will prove far more than mere existence.
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Proof.

For \(x=0\text{,}\) the existence of a unique root is trivial.
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Let \(f \colon (0,\infty) \to (0,\infty)\) be defined by \(f(y) \coloneqq y^n\text{.}\) The function \(f\) is continuously differentiable, and \(f'(y) = ny^{n-1}\text{,}\) see Exercise 4.1.3. For \(y > 0\text{,}\) the derivative \(f'\) is strictly positive, and so again by Proposition 4.2.8, \(f\) is strictly increasing (this can also be proved directly) and hence injective. Suppose \(M\) and \(\epsilon\) are such that \(M > 1\) and \(1 > \epsilon > 0\text{.}\) Then \(f(M) = M^n \geq M\) and \(f(\epsilon) = \epsilon^n \leq \epsilon\text{.}\) For every \(x\) with \(\epsilon < x < M\text{,}\) we have, by the intermediate value theorem, that \(x \in f\bigl( [\epsilon,M] \bigr) \subset f\bigl( (0,\infty) \bigr)\text{.}\) As \(M\) and \(\epsilon\) were arbitrary, \(f\) is onto \((0,\infty)\text{,}\) and hence \(f\) is bijective. Let \(g\) be the inverse of \(f\text{,}\) and we obtain the existence and uniqueness of positive \(n\)th roots. Lemma 4.4.1 says that \(g\) has a continuous derivative and \(g'(x) = \frac{1}{f'\left(\vphantom{1^1_1}g(x)\right)} = \frac{1}{n {(x^{1/n})}^{n-1}}\text{.}\)
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Example 4.4.4.

The corollary provides a good example of where the inverse function theorem gives us an interval smaller than \((a,b)\text{.}\) Take \(f \colon \R \to \R\) defined by \(f(x) \coloneqq x^2\text{.}\) Then \(f'(x_0) \neq 0\) as long as \(x_0 \neq 0\text{.}\) If \(x_0 > 0\text{,}\) we can take \(I=(0,\infty)\text{,}\) but no larger.
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Example 4.4.5.

Another useful example is \(f(x) \coloneqq x^3\text{.}\) The function \(f \colon \R \to \R\) is one-to-one and onto, so \(f^{-1}(y) = y^{1/3}\) exists on the entire real line, including zero and negative \(y\text{.}\) The function \(f\) has a continuous derivative, but \(f^{-1}\) has no derivative at the origin. The point is that \(f'(0) = 0\text{.}\) See Figure 4.11 for a graph. Notice the vertical tangent on the cube root at the origin. See also Exercise 4.4.4.
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The graph y equals the cube root of x in bold line and the graph of y equals x cubed in dashed line. We note that while the graph of x cubed is tangent to the x-axis at the origin, the cube root is a reflection across the x equals y line and is therefore tangent to the vertical y-axis at the origin.
Figure 4.11. Graphs of \(x^3\) and \(x^{1/3}\text{.}\)
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Exercises 4.4.2 Exercises

4.4.1.

Suppose \(f \colon \R \to \R\) is continuously differentiable and \(f'(x) > 0\) for all \(x\text{.}\) Show that \(f\) is invertible on the interval \(J = f(\R)\text{,}\) the inverse is continuously differentiable, and \({(f^{-1})}'(y) > 0\) for all \(y \in f(\R)\text{.}\)
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4.4.2.

Suppose \(I,J\) are intervals and a monotone onto \(f \colon I \to J\) has an inverse \(g \colon J \to I\text{.}\) Suppose you already know that both \(f\) and \(g\) are differentiable everywhere and \(f'\) is never zero. Using the chain rule but not Lemma 4.4.1, prove the formula \(g'(y) = \frac{1}{f'\left(\vphantom{1_1^1}g(y)\right)}\text{.}\) Remark: This exercise is the same as Exercise 4.1.10, no need to do it again if you have solved it already.
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4.4.3.

Let \(n\in \N\) be even. Prove that every \(x > 0\) has a unique negative \(n\)th root. That is, there exists a negative number \(y\) such that \(y^n = x\text{.}\) Compute the derivative of the function \(g(x) \coloneqq y\text{.}\)
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4.4.4.

Let \(n \in \N\) be odd and \(n \geq 3\text{.}\) Prove that every \(x\) has a unique \(n\)th root. That is, there exists a number \(y\) such that \(y^n = x\text{.}\) Prove that the function defined by \(g(x) \coloneqq y\) is differentiable except at \(x=0\) and compute the derivative. Prove that \(g\) is not differentiable at \(x=0\text{.}\)
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4.4.5.

(requires Section 4.3)   Show that if in the inverse function theorem \(f\) has \(k\) continuous derivatives, then the inverse function \(g\) also has \(k\) continuous derivatives.
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4.4.6.

Let \(f(x) \coloneqq x + 2 x^2 \sin(\nicefrac{1}{x})\) for \(x \neq 0\) and \(f(0) \coloneqq 0\text{.}\) Show that \(f\) is differentiable at all \(x\text{,}\) that \(f'(0) > 0\text{,}\) but that \(f\) is not invertible on any open interval containing the origin.
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4.4.7.

  1. Let \(f \colon \R \to \R\) be a continuously differentiable function and \(k > 0\) be a number such that \(f'(x) \geq k\) for all \(x \in \R\text{.}\) Show \(f\) is one-to-one and onto, and has a continuously differentiable inverse \(f^{-1} \colon \R \to \R\text{.}\)
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  2. Find an example \(f \colon \R \to \R\) where \(f'(x) > 0\) for all \(x\text{,}\) but \(f\) is not onto.
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4.4.8.

Suppose \(I,J\) are intervals and a monotone onto \(f \colon I \to J\) has an inverse \(g \colon J \to I\text{.}\) Suppose \(x \in I\) and \(y \coloneqq f(x) \in J\text{,}\) and that \(g\) is differentiable at \(y\text{.}\) Prove:
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  1. If \(g'(y) \neq 0\text{,}\) then \(f\) is differentiable at \(x\text{.}\)
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  2. If \(g'(y) = 0\text{,}\) then \(f\) is not differentiable at \(x\text{.}\)
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