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Our choice ensures that \(f(a-\epsilon)\leq f(a)-\delta\) and \(f(a)+\delta\leq f(a+\epsilon)\); therefore \(|x-f(a)|<\delta\) implies \(f(a-\epsilon) \[f^{-1}\left(f(a-\epsilon)\right) or equivalently \(|f^{-1}(x)-a|<\epsilon\), which is what we wanted to show. Now that we established the continuity of \(f^{-1}\), the next step is to investigate the differentiability of \(f^{-1}\). Again the following picture gives us the hope that tangent line \(L\) at the point \((a,f(a))\) to a one-to-one function \(f\) should be related nicely to the tangent line \(L^{*}\) at the point \((f(a),a)\) to the function \(f^{-1}\) (Figure 1.44). Figure 1.43: Continuity of \(f^{-1}\)If we rush and apply the Chain Rule to \(f\left(f^{-1}(x)\right)=x\) by differentiating both sides, we get \[f^{\prime}\left(f^{-1}(x)\right)(f^{-1})^{\prime}(x)=1,\] so \[(f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)}.\] But this is not a proof that \(f^{-1}\) is differentiable; we cannot even apply the Chain Rule unless \(f^{-1}\) is already known to be differentiable. But suppose that \(f^{-1}\) is differentiable, then from above we know what the derivative \((f^{-1})^{\prime}(x)\) is going to look like. This observation also teaches us that even if \(f\) is a continuous one-to-one function defined on an interval and if \(f^{\prime}\left(f^{-1}(a)\right)=0\), then \(f^{-1}\) is not differentiable at \(a\). Let \(y=f(x)\) be a \(\mathcal{C}^{1}\) -function and \(f^{\prime}(a)\neq 0\); then the following theorem asserts that locally near \(a\) we can solve for \(x\) to give an inverse function \(x=f^{-1}(y)\). We also learn that since \(f^{\prime}(y)=\frac{1}{f^{\prime}(x)}\), it is possible that \(y=f(x)\) can be inverted because \(f^{\prime}(a)\neq 0\) means that the slope of \(y=f(x)\) is nonzero, so that the graph is rising or falling near \(a\). Thus if we reflect the graph across the line \(y=x\), it is still a graph near \((a,b)\), where \(b=f(a)\). In the following figure we can invert \(y=f(x)\) in the shaded box, so that in this range \(x=f^{-1}(y)\) is defined. In summary, if \(f^{\prime}(a)\neq 0\), then \(y=f(x)\) is locally invertible (Figure 1.45). Now we are ready to state the Figure 1.45: Local invertibility **Theorem 49** (Inverse Function Theorem).: _Let \(f\) be a continuous one-to-one function defined on an interval, and suppose that \(f\) is differentiable at \(f^{-1}(b)\), with derivative \(f^{\prime}\left(f^{-1}(b)\right)\neq 0\). Then \(f^{-1}\) is differentiable at \(b\), and_ \[(f^{-1})^{\prime}(b)=\frac{1}{f^{\prime}\left(f^{-1}(b)\right)}.\] : Since we have a continuous one-to-one function \(f\), by the above theorems we deduce that \(f\) is monotone, and \(f^{-1}\) exists and is continuous. Let \(b=f(a)\), then \(f\) differentiable at \(f^{-1}(b)\) ensures that the limit \[f^{\prime}(a)=\lim_{k\to 0}\frac{f(a+k)-f(a)}{k}\] exists. Moreover, by hypothesis, \(f^{\prime}(a)\neq 0\). Now consider the limit \[\lim_{h\to 0}\frac{f^{-1}(b+h)-f^{-1}(b)}{h}.\] Since \(b+h\) is a number in the domain of \(f^{-1}\), it can be written as \(f(a+k)\) for some \(k=k(h)\), and rewriting the limit \[\lim_{h\to 0}\frac{f^{-1}(b+h)-f^{-1}(b)}{h}=\lim_{h\to 0}\frac{f^{-1}(f(a+k))-a}{h}= \lim_{k\to 0}\frac{k}{f(a+k)-b}\] provided that \(k\to 0\) as \(h\to 0\) too. This is not difficult to show and follows from the continuity of \(f^{-1}\) at \(b\), i.e., \(\lim_{h\to 0}f^{-1}(b+h)=f^{-1}(b)\). Now we set \[b+h=f(a+k),\quad\text{thus}\quad k=f^{-1}(b+h)-f^{-1}(b).\] Since \[\lim_{k\to 0}\frac{f(a+k)-f(a)}{k}=f^{\prime}(a)=f^{\prime}(f^{-1}(b))\neq 0,\] we have that \[(f^{-1})^{\prime}(b)=\frac{1}{f^{\prime}\left(f^{-1}(b)\right)}.\qed\] **Example 48**.: Note that \(f(2)=6\) and \(f^{\prime}(x)=2x\), so \(f^{\prime}(2)=4\). By the Inverse Function Theorem, \[(f^{-1})^{\prime}(6)=\ \frac{1}{f^{\prime}(2)}=\frac{1}{4}.\] We can check this by finding \((f^{-1})^{\prime}(x)=\frac{1}{2}(x-2)^{-1/2}\), so \((f^{-1})^{\prime}(6)=1/4\). **Remark 27**.: There is a wonderful calculus book by Spivak [46], pp. 200-210, which contains a detailed discussion of inverse functions. See also Section 1.6 for continuous functions on compact sets and Section 2.1 for metric spaces. It turns out that if \(f\) is a continuous and one-to-one mapping of a compact metric space \(M\) onto a metric space \(N\), then the inverse mapping \(f^{-1}\) can be defined on \(N\) by \(f^{-1}(f(x))=x\) for all \(x\in M\) and \(f^{-1}\) is a continuous mapping of \(N\) onto \(M\) (see, e.g., [42], p. 221). For the Inverse Function Theorem for a \(\mathcal{C}^{1}\)-function \(f:A\subset\mathbb{R}^{n}\to\mathbb{R}^{n}\) (a function of several variables), we refer the reader to [26], p. 392. ### Exercises 1. Discuss the differentiability of \(f(x)=|x^{2}-9|\) at \(x=2\). 3. Let \(f\) be a function on \([a,b]\) that is differentiable at \(c\). Let \(L(x)\) be the tangent line to \(f\) at \(c\). Prove that \(L\) is the unique linear function with the property that \[\lim_{x\to c}\frac{f(x)-L(x)}{x-c}=0.\] 4. Find the derivative (if it exists) of \[f(x)=\left\{\begin{array}{ll}x^{2}e^{-x^{2}}&\qquad\mbox{if }|x|\leq 1\\ \frac{1}{e}&\qquad\mbox{if }|x|>1.\end{array}\right.\] 5. We say a function \(f:(a,b)\to\mathbb{R}\) is uniformly differentiable if \(f\) is differentiable on \((a,b)\), and for each \(\epsilon>0\) there exists a \(\delta>0\) such that \[0<|x-y|<\delta\quad\mbox{ and }\quad x,y\in(a,b)\quad\Rightarrow\quad\left| \frac{f(x)-f(y)}{x-y}-f^{\prime}(x)\right|<\epsilon.\] Prove that if \(f\) is uniformly differentiable, then \(f^{\prime}\) is continuous. Then give an example of a function that is differentiable but not uniformly differentiable. 6. Let \(f:\mathbb{R}\to\mathbb{R}\). Assume that for any \(x,t\in\mathbb{R}\), we have \[|f(x)-f(t)|\leq|x-t|^{1+\alpha},\] where \(\alpha>0\). Show that \(f(x)\) is constant. 7. Does the Mean Value Theorem apply to \(f(x)=\sqrt{x}\) on \([0,1]\)? Does it apply to \(g(x)=\sqrt{|x|}\) on \([-1,1]\)? 8. Consider the function \[f(x)=\left\{\begin{array}{ll}e^{-1/x^{2}}&\mbox{if }x\neq 0\\ &\\ 0&\mbox{if }x=0.\end{array}\right.\] Show that the \(n\)th derivative \(f^{(n)}\) at \(0\) is \(f^{(n)}(0)=0\), for \(n=1,2,\dots\). 9. Suppose \(f\) and \(g\) are real functions of a real variable whose \(n\)th derivatives \(f^{(n)}\) exist at a point \(a\), where \(n\in\mathbb{N}\). Prove the Leibniz generalization of the product formula: \[(fg)^{(n)}(a)=\sum_{k=0}^{n}\binom{n}{k}f^{(k)}(a)g^{(n-k)}(a).\]10. For \(x\in\mathbb{R}\), let \(f\) be defined as \(f(x)=x^{2}e^{x^{2}}\). Show that \(f^{-1}\) exists and is differentiable on \((0,\infty)\). Compute its derivative \((f^{-1})^{\prime}(e)\). 11. Suppose \(f\) and \(g\) are one-to-one continuous functions on \(\mathbb{R}\). Compute the derivatives: 1. \((f^{-1})^{\prime}(2)\)
Proof
1. Let \(f(x)=x^{3}\). Since \(f^{\prime}(x)=3x^{2}\) and thus \(f^{\prime}(0)=0\), the inverse function \(f^{-1}(x)=\sqrt[3]{x}\) is not differentiable at \(0\). Note that this example illustrates the necessity of the condition \(f^{\prime}\left(f^{-1}(b)\right)\neq 0\) in the above theorem.
2. Let \(f:[0,\infty)\to\mathbb{R}\) be defined as \(f(x)=x^{2}+2\). Then the function \(f^{-1}(x):[2,\infty)\to\mathbb{R}\) defined by \(f^{-1}(x)=\sqrt{x-2}\) is the inverse function, since \[f(f^{-1}(x))=\left(\sqrt{x-2}\right)^{2}+2=x-2+2=x\quad\text{for all }x\in[2,\infty)\]\[f^{-1}(f(x))=\sqrt{(x^{2}+2)-2}=\sqrt{x^{2}}=x\quad\text{for all}\ \ x\in[0,\infty).\]
3. For \(n\) odd let \(f(x)=x^{n}\) for all \(x\), and for \(n\) even, let \(f(x)=x^{n}\) for \(x\geq 0\). Then \(f\) is a continuous and one-to-one function whose inverse function is \(f^{-1}(x)=\sqrt[n]{x}\). By the Inverse Function Theorem, we have for \(x\neq 0\) \[(f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)}=\frac{1}{n(x^ {1/n})^{n-1}}=\frac{1}{n}x^{(1/n)-1}.\] What we have shown is if one considers a function \(f(x)=x^{a}\), where \(a\) is an integer or the reciprocal of a natural number, then the derivative is given by \(f^{\prime}(x)=ax^{a-1}\). Now you can check that this formula is also true if \(a\) is a rational number of the form \(a=\frac{m}{n}\).
2. Determine whether or not \(f\) is differentiable at \(0\). 1. \(f(x)=\sqrt[3]{x}\). 2. \(f(x)=\sqrt{|x|}\). 3. \(f:\mathbb{R}\to\mathbb{R}\) defined by \(f(x)=x\sin\left(\frac{1}{x}\right)\) if \(x\neq 0\) and \(f(0)=0\). 4. \(f(x)=x|x|\).
2. \((g^{-1})^{\prime}(2)\)
3. \((f^{-1}\cdot g^{-1})^{{}^{\prime}}(2)\)
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