Everyone who has studied calculus believes that a function whose derivative is zero at every point of an interval is constant. Try to prove it from the definition of the derivative. The hypothesis says that at each point \(a\) the difference quotient tends to \(0\) : near \(a\) , the function changes by much less than the distance moved. That is a statement about each point separately, with its own “near”. The conclusion compares the values of the function at two points that may be far apart. Between them lie infinitely many points, each with its own tiny neighbourhood and its own tiny error, and nothing in the definition prevents infinitely many tiny errors from adding up.

The gap between the local definition and global conclusions is bridged by one theorem, the Mean Value Theorem, and it is not a formal consequence of the definition: it uses the Extreme Value Theorem, hence compactness, hence completeness. This chapter proves it and then uses it for everything else, including monotonicity, inequalities, L’Hôpital’s rule, and Taylor’s theorem. Along the way it shows that derivatives, though they need not be continuous, cannot be arbitrary functions.

6.1 The derivative

Throughout the chapter \(I\) denotes an interval containing more than one point.

The limit is taken over \(x\in I\) , \(x\ne a\) , so at an endpoint of \(I\) the derivative is automatically one-sided. A reformulation due to Carathéodory removes the quotient and makes most proofs shorter.

The Carathéodory form disposes of the classical difficulty in the chain rule, that \(f(x)-f(a)\) may vanish for \(x\) near \(a\) : no division by it occurs. From (b) and induction, \((x^n)'=nx^{n-1}\) , and with (a) and (c) every polynomial and rational function is differentiable on its domain. The resulting table of derivatives is the one built up in Calculus, where these rules are first met as computational structure (math.calculus.derivative-rules-as-structure). The present chapter is about what the derivative implies, and uses that table freely.

Example 6.2: A derivative that exists everywhere and is not continuous

Let \(f(x)=x^2\sin(1/x)\) for \(x\ne 0\) and \(f(0)=0\) . For \(x\ne 0\) the rules give

\[ f'(x)=2x\sin(1/x)-\cos(1/x). \]

At \(0\) the rules do not apply and we use the definition: the quotient is \(f(x)/x=x\sin(1/x)\) , and \(\lvert x\sin(1/x)\rvert\le\lvert x\rvert\to 0\) . So \(f'(0)=0\) . The function is differentiable at every point of \(\mathbb R\) .

But \(f'\) is not continuous at \(0\) . At the points \(x_n=1/(2\pi n)\) , which tend to \(0\) , we have \(\sin(1/x_n)=0\) and \(\cos(1/x_n)=1\) , so \(f'(x_n)=-1\) for every \(n\) , while \(f'(0)=0\) . At the points \(1/((2n+1)\pi)\) the derivative equals \(+1\) . The derivative oscillates between values near \(-1\) and near \(1\) in every neighbourhood of \(0\) .

So “differentiable” and “continuously differentiable” are different properties. A function whose derivative exists and is continuous on \(I\) is said to be of class \(C^1\) .

6.2 From local to global: the Mean Value Theorem

A point \(c\in I\) is a local maximum of \(f\) if there is \(\delta>0\) with \(f(x)\le f(c)\) for all \(x\in I\) with \(\lvert x-c\rvert<\delta\) ; local minimum is defined likewise. An interior point of \(I\) is a point that is not an endpoint.

The theorem asserts that a point \(c\) exists; it does not say where it is or that it is unique, and in applications one almost never finds it. Its use is to transfer a bound on \(f'\) over an interval into a bound on a difference of values of \(f\) .

6.3 What a derivative can and cannot be

Example 6.2 showed that a derivative need not be continuous. One might then expect that any function at all could be a derivative. It cannot: derivatives share with continuous functions the property of not skipping values.

Note that the proof does not assume \(f'\) continuous; if it did, the Intermediate Value Theorem would apply directly. A consequence is that a derivative has no jump discontinuities. The function equal to \(0\) for \(x<0\) and \(1\) for \(x\ge 0\) skips the value \(\tfrac12\) and so is not the derivative of any function on \(\mathbb R\) . The discontinuity in Example 6.2 is of the only kind a derivative can have: oscillation.

6.4 Cauchy’s mean value theorem and L’Hôpital’s rule

With \(g(x)=x\) this is Theorem 6.7. The point is that the same \(c\) serves both functions, which two separate applications of the Mean Value Theorem would not give.

There is also a version for quotients in which \(g(x)\to\infty\) ; it is proved from the same theorem with one more estimate (Exercise 21 outlines it).

6.5 Taylor’s theorem

The Mean Value Theorem writes \(f(x)=f(a)+f'(c)(x-a)\) : the constant approximation \(f(a)\) , plus an error expressed through the first derivative at an unknown point. Taylor’s theorem does the same for polynomial approximations of any degree. For a function \(f\) with \(n\) derivatives at \(a\) , its Taylor polynomial of degree \(n\) at \(a\) is

It is the unique polynomial of degree at most \(n\) whose value and first \(n\) derivatives at \(a\) agree with those of \(f\) .

For \(n=0\) this is the Mean Value Theorem. The remainder term has the form of the next term of the polynomial, with the derivative evaluated at an unknown intermediate point; a bound on \(f^{(n+1)}\) between \(a\) and \(x\) therefore bounds the error.

6.6 Derivatives of inverse functions

The hypothesis \(f'(a)\ne 0\) cannot be removed: \(f(x)=x^3\) is strictly increasing with \(f'(0)=0\) , and its inverse \(y^{1/3}\) is not differentiable at \(0\) . For \(f(x)=x^n\) on \((0,\infty)\) the theorem gives the derivative of \(g(y)=y^{1/n}\) as \(1/(n\,g(y)^{n-1})=\tfrac1n\,y^{1/n-1}\) , extending the power rule to roots.

The chapter has gone from the definition of the derivative to global conclusions in one essential step, the Mean Value Theorem. The reverse journey, from a derivative back to the function, is integration. Chapter 7 defines the integral without reference to derivatives and then proves that the two operations undo each other; the proof of that fact will use the Mean Value Theorem once more.

Exercises

1. Compute from Definition 6.1, by finding the function \(\varphi\) of Proposition 6.2 explicitly, the derivative of \(f(x)=1/x\) at a point \(a\ne 0\) and of \(f(x)=\sqrt x\) at a point \(a>0\) .

   Answer

   For \(f(x)=1/x\) : \(\dfrac{1/x-1/a}{x-a}=-\dfrac{1}{xa}\) , so \(\varphi(x)=-1/(xa)\) , continuous at \(a\) , and \(f'(a)=-1/a^2\) . For \(f(x)=\sqrt x\) : \(\varphi(x)=1/(\sqrt x+\sqrt a)\) , and \(f'(a)=1/(2\sqrt a)\) .

2. Let \(f(x)=x\lvert x\rvert\) . Show that \(f\) is differentiable on \(\mathbb R\) , find \(f'\) , and decide whether \(f'\) is differentiable at \(0\) .

   Solution

   Here \(f(x)=x^2\) for \(x\ge 0\) and \(f(x)=-x^2\) for \(x<0\) .

   Away from \(0\) . If \(a>0\) , then \(f(x)=x^2\) for all \(x\) in a neighbourhood of \(a\) , and the derivative at a point depends only on the function near that point, so \(f'(a)=2a\) . If \(a<0\) , similarly \(f'(a)=-2a\) . In both cases \(f'(a)=2\lvert a\rvert\) .

   At \(0\) . The formulas on the two sides differ, so we use the definition. For \(x\ne 0\) ,

   \[ \frac{f(x)-f(0)}{x-0}=\frac{x\lvert x\rvert}{x}=\lvert x\rvert\to 0\quad\text{as }x\to 0 . \]

   So \(f'(0)=0\) , which also equals \(2\lvert 0\rvert\) .

   Hence \(f\) is differentiable on \(\mathbb R\) with \(f'(x)=2\lvert x\rvert\) , a continuous function: \(f\) is of class \(C^1\) .

   Second derivative at \(0\) . The difference quotient of \(f'\) at \(0\) is \(2\lvert x\rvert/x\) , which is \(2\) for \(x>0\) and \(-2\) for \(x<0\) . It has no limit, so \(f'\) is not differentiable at \(0\) . The function \(f\) is once but not twice differentiable at \(0\) ; multiplying \(\lvert x\rvert\) by \(x\) smoothed the corner by exactly one order.

3. Let \(f(x)=x^2\) for rational \(x\) and \(f(x)=0\) for irrational \(x\) . Prove that \(f\) is differentiable at \(0\) , and that \(f\) is not continuous, hence not differentiable, at any other point.

   Solution

   At \(0\) . For \(x\ne 0\) the difference quotient is \(f(x)/x\) , which equals \(x\) if \(x\) is rational and \(0\) if \(x\) is irrational. In both cases \(\lvert f(x)/x\rvert\le\lvert x\rvert\) . Given \(\varepsilon>0\) , take \(\delta=\varepsilon\) : if \(0<\lvert x\rvert<\delta\) then \(\lvert f(x)/x-0\rvert<\varepsilon\) . So \(f'(0)=0\) .

   At \(a\ne 0\) . By the density of the rationals and of the irrationals (Theorem 1.10 and Corollary 1.12) there are rationals \(q_n\to a\) and irrationals \(t_n\to a\) . Then

   \[ f(q_n)=q_n^2\to a^2\ne 0,\qquad f(t_n)=0\to 0 . \]

   If \(f\) were continuous at \(a\) , both sequences \(f(q_n)\) and \(f(t_n)\) would converge to \(f(a)\) by Theorem 5.2, forcing \(a^2=0\) . So \(f\) is not continuous at \(a\) , and by Corollary 6.3 it is not differentiable at \(a\) .

   Thus \(f\) is differentiable at exactly one point, and continuous at exactly one point. Differentiability at a point is a statement about that point only; it implies nothing about the function’s behaviour at any other point, however near.

4. Prove the quotient rule of Theorem 6.4(c) using Proposition 6.2, first for the reciprocal \(1/g\) and then by combining with the product rule.

   Hint

   Since \(g\) is continuous at \(a\) and \(g(a)\ne 0\) , \(g(x)\ne 0\) for \(x\) near \(a\) (Exercise 5.3). For such \(x\) , \(\dfrac1{g(x)}-\dfrac1{g(a)}=-\dfrac{g(x)-g(a)}{g(x)g(a)}=-\dfrac{\psi(x)}{g(x)g(a)}\,(x-a)\) , and the coefficient is continuous at \(a\) with value \(-g'(a)/g(a)^2\) . Then apply the product rule to \(f\cdot(1/g)\) .

5. For each integer \(k\ge 0\) let \(f_k(x)=x^k\sin(1/x)\) for \(x\ne 0\) and \(f_k(0)=0\) . For which \(k\) is \(f_k\) continuous at \(0\) ? Differentiable at \(0\) ? Continuously differentiable at \(0\) ?

   Answer

   Continuous at \(0\) exactly for \(k\ge 1\) (for \(k=0\) , \(\sin(1/x)\) has no limit). Differentiable at \(0\) exactly for \(k\ge 2\) , since the difference quotient is \(x^{k-1}\sin(1/x)\) . Continuously differentiable at \(0\) exactly for \(k\ge 3\) : for \(x\ne 0\) , \(f_k'(x)=kx^{k-1}\sin(1/x)-x^{k-2}\cos(1/x)\) , which tends to \(f_k'(0)=0\) only if \(k-2\ge 1\) .

6. Find all points \(c\) that satisfy the conclusion of the Mean Value Theorem for \(f(x)=x^3-x\) on \([0,2]\) and for \(f(x)=1/x\) on \([1,4]\) .

   Answer

   For \(x^3-x\) on \([0,2]\) : the secant slope is \((6-0)/2=3\) , and \(3c^2-1=3\) gives \(c=2/\sqrt3\approx 1.1547\) (the negative root is outside the interval). For \(1/x\) on \([1,4]\) : the secant slope is \((\tfrac14-1)/3=-\tfrac14\) , and \(-1/c^2=-\tfrac14\) gives \(c=2\) .

7. Use the Mean Value Theorem to prove that \(x/(1+x)<\log(1+x)<x\) for every \(x>0\) .

   Solution

   Fix \(x>0\) and let \(f(t)=\log(1+t)\) for \(t\in[0,x]\) . It is continuous on \([0,x]\) and differentiable on \((0,x)\) with \(f'(t)=1/(1+t)\) . By the Mean Value Theorem there is \(c\in(0,x)\) with

   \[ \log(1+x)-\log 1=\frac{1}{1+c}\,(x-0),\qquad\text{that is,}\qquad\log(1+x)=\frac{x}{1+c}. \]

   Since \(0<c<x\) we have \(1<1+c<1+x\) , so \(\dfrac{1}{1+x}<\dfrac{1}{1+c}<1\) . Multiplying by \(x>0\) ,

   \[ \frac{x}{1+x}<\log(1+x)<x . \]

   For example, with \(x=0.5\) the bounds are \(0.3333<\log 1.5<0.5\) , and \(\log 1.5\approx 0.4055\) . Neither the value of \(c\) nor its uniqueness was needed, only the interval in which it lies.

8. Let \(r\ge 1\) be a real number. Use the Mean Value Theorem to prove that \((1+x)^r\ge 1+rx\) for all \(x>-1\) , extending Lemma 2.6 from integer to real exponents.

   Hint

   Let \(f(t)=(1+t)^r-1-rt\) on \((-1,\infty)\) . Then \(f(0)=0\) and \(f'(t)=r\bigl((1+t)^{r-1}-1\bigr)\) , which has the sign of \(t\) when \(r>1\) and is identically \(0\) when \(r=1\) . So \(f\) is decreasing on \((-1,0]\) and increasing on \([0,\infty)\) by Corollary 6.8(b), and its minimum value is \(f(0)=0\) .

9. Let \(f:\mathbb R\to\mathbb R\) be differentiable with \(\lvert f'(x)\rvert\le c\) for all \(x\) , where \(c<1\) . Prove that for any starting value \(x_1\) the sequence \(x_{n+1}=f(x_n)\) converges, and that its limit is the unique solution of \(f(x)=x\) .

   Solution

   Contraction. By Corollary 6.8(c), \(\lvert f(x)-f(y)\rvert\le c\lvert x-y\rvert\) for all real \(x,y\) . Hence

   \[ \lvert x_{n+2}-x_{n+1}\rvert=\lvert f(x_{n+1})-f(x_n)\rvert\le c\,\lvert x_{n+1}-x_n\rvert . \]

   If \(c>0\) the sequence is contractive in the sense of Proposition 2.17; if \(c=0\) the same inequality holds with any constant in \((0,1)\) . Either way \((x_n)\) converges; call the limit \(L\) .

   The limit is a fixed point. \(f\) is differentiable, hence continuous, so \(f(x_n)\to f(L)\) . But \(f(x_n)=x_{n+1}\to L\) . By uniqueness of limits, \(f(L)=L\) .

   Uniqueness. If \(f(L)=L\) and \(f(L')=L'\) , then \(\lvert L-L'\rvert=\lvert f(L)-f(L')\rvert\le c\lvert L-L'\rvert\) , so \((1-c)\lvert L-L'\rvert\le 0\) . Since \(1-c>0\) , \(L=L'\) .

   For instance \(f(x)=\tfrac12\cos x\) has \(\lvert f'\rvert\le\tfrac12\) . Starting from \(x_1=1\) the iteration gives \(0.2702\) , \(0.4819\) , \(0.4431\) , \(0.4517\) , and converges to the unique solution of \(\cos x=2x\) , which is \(0.450184\) to six places.

10. Find the error: “For \(f(x)=1/x\) we have \(f'(x)=-1/x^2<0\) at every point of the domain, so \(f\) is decreasing; hence \(f(-1)>f(1)\) , that is, \(-1>1\) .”

    Answer

    Corollary 6.8(b) is a statement about an interval. The domain of \(1/x\) is \((-\infty,0)\cup(0,\infty)\) , which is not an interval. The function is decreasing on each of the two pieces separately, and the corollary says nothing about how values on one piece compare with values on the other.

11. Let \(f\) be continuous on \([a,b]\) and differentiable on \((a,b)\) with \(f'(x)>0\) except at finitely many points, where \(f'(x)=0\) . Prove that \(f\) is strictly increasing on \([a,b]\) .

    Hint

    By Corollary 6.8(b), \(f\) is increasing. If \(f(x)=f(y)\) for some \(x<y\) , then \(f\) is constant on \([x,y]\) , so \(f'=0\) at every point of \((x,y)\) , which is infinitely many points.

12. Prove that a differentiable function on an interval with bounded derivative is uniformly continuous. Show that the converse is false by considering \(\sqrt x\) on \((0,1)\) .

    Hint

    Corollary 6.8(c) gives a Lipschitz bound, and Lipschitz functions are uniformly continuous. For the converse, \(\sqrt x\) on \((0,1)\) extends continuously to \([0,1]\) , so it is uniformly continuous by Theorem 5.15(b), but its derivative \(1/(2\sqrt x)\) is unbounded.

13. Prove that there is no differentiable function \(F:\mathbb R\to\mathbb R\) with \(F'(x)=0\) for \(x<0\) and \(F'(x)=1\) for \(x\ge 0\) .

    Solution

    Suppose such an \(F\) exists. It is differentiable on \([-1,1]\) , with \(F'(-1)=0\) and \(F'(1)=1\) . The number \(\lambda=\tfrac12\) lies strictly between these values, so by Darboux’s theorem (Theorem 6.9) there is \(c\in(-1,1)\) with \(F'(c)=\tfrac12\) . But \(F'\) takes only the values \(0\) and \(1\) . This is a contradiction, so no such \(F\) exists.

    A direct argument shows where the trouble is. By Corollary 6.8(a) applied on \((-\infty,0]\) , \(F\) would be constant there, say \(F(x)=k\) . On \([0,\infty)\) the function \(F(x)-x\) has derivative \(0\) , so \(F(x)=x+k\) , the constant being the same because \(F(0)=k\) . Then the difference quotient of \(F\) at \(0\) is \(0\) from the left and \(1\) from the right, so \(F'(0)\) does not exist, contrary to the requirement \(F'(0)=1\) . The candidate antiderivative has a corner exactly where the proposed derivative jumps.

14. Let \(f:\mathbb R\to\mathbb R\) be differentiable with \(f'(x)\ne 0\) for every \(x\) . Prove that \(f\) is strictly monotone.

    Solution

    Suppose \(f'\) takes a positive value at some point \(p\) and a negative value at some point \(q\) . Then \(p\ne q\) , and \(0\) lies strictly between \(f'(q)\) and \(f'(p)\) . Applying Darboux’s theorem on the closed interval with endpoints \(p\) and \(q\) , on which \(f\) is differentiable, there is a point \(c\) between them with \(f'(c)=0\) . This contradicts the hypothesis.

    Therefore \(f'\) does not change sign: since it is never \(0\) , either \(f'(x)>0\) for all \(x\) or \(f'(x)<0\) for all \(x\) . By Corollary 6.8(b), \(f\) is strictly increasing in the first case and strictly decreasing in the second.

    Note that \(f'\) was not assumed continuous. Had it been, the Intermediate Value Theorem would give the zero directly; Darboux’s theorem is what allows the argument for every differentiable function.

15. Evaluate with L’Hôpital’s rule, checking the hypotheses each time: (a) \(\lim_{x\to 0}(e^x-1-x)/x^2\) ; (b) \(\lim_{x\to 1}\log x/(x-1)\) ; (c) \(\lim_{x\to 0}(x-\sin x)/x^3\) .

    Answer

    (a) \(\tfrac12\) : both limits are \(0\) ; the quotient of derivatives is \((e^x-1)/(2x)\) , again of the same type, with quotient of derivatives \(e^x/2\to\tfrac12\) . (b) \(1\) : both tend to \(0\) ; the quotient of derivatives is \((1/x)/1\to 1\) . (c) \(\tfrac16\) : three applications, through \((1-\cos x)/(3x^2)\) and \(\sin x/(6x)\) to \(\cos x/6\to\tfrac16\) .

16. A student computes \(\lim_{x\to 0}(x+1)/(2x+3)\) “by L’Hôpital’s rule” as \(\lim_{x\to 0}1/2=\tfrac12\) . Find the correct limit and the hypothesis that was not checked.

    Answer

    The function is continuous at \(0\) , so the limit is its value there, \(\tfrac13\) . L’Hôpital’s rule requires the numerator and the denominator both to tend to \(0\) ; here they tend to \(1\) and \(3\) .

17. Let \(f(x)=\sqrt{1+x}\) . Write down its Taylor polynomial \(P_2\) at \(0\) , use it to approximate \(\sqrt{1.1}\) , and use Theorem 6.12 to bound the error.

    Solution

    Derivatives. With \(f(x)=(1+x)^{1/2}\) on \((-1,\infty)\) ,

    \[ f'(x)=\tfrac12(1+x)^{-1/2},\qquad f''(x)=-\tfrac14(1+x)^{-3/2},\qquad f'''(x)=\tfrac38(1+x)^{-5/2}. \]

    So \(f(0)=1\) , \(f'(0)=\tfrac12\) , \(f''(0)=-\tfrac14\) , and

    \[ P_2(x)=1+\frac x2-\frac{x^2}{8}. \]

    Approximation. \(P_2(0.1)=1+0.05-0.00125=1.04875\) .

    Error. By Theorem 6.12 with \(n=2\) , \(a=0\) , \(x=0.1\) ,

    \[ \sqrt{1.1}-P_2(0.1)=\frac{f'''(c)}{3!}\,(0.1)^3\qquad\text{for some }c\in(0,0.1). \]

    For such \(c\) , \(0<f'''(c)=\tfrac38(1+c)^{-5/2}<\tfrac38\) . Hence

    \[ 0<\sqrt{1.1}-1.04875<\frac{3/8}{6}\cdot 0.001=0.0000625 . \]

    So \(1.04875<\sqrt{1.1}<1.0488125\) . The true value is \(1.0488088\) , with error about \(0.0000588\) , inside the bound and of the predicted sign.

    Two features of the estimate are worth noticing. Because the third derivative is positive on the whole interval, the theorem gives not only the size of the error but its direction: the quadratic polynomial underestimates the square root. And the bound was obtained by replacing the unknown point \(c\) with the endpoint at which the third derivative is largest, which is the usual way to make the remainder computable.

18. Let \(f\) be twice differentiable on an open interval \(I\) with \(f''(x)\ge 0\) for all \(x\in I\) . Prove that the graph of \(f\) lies above each of its tangent lines: \(f(x)\ge f(a)+f'(a)(x-a)\) for all \(a,x\in I\) .

    Solution

    Let \(a,x\in I\) . If \(x=a\) the inequality is an equality. If \(x\ne a\) , apply Taylor’s theorem (Theorem 6.12) with \(n=1\) : there is a point \(c\) strictly between \(a\) and \(x\) , hence in \(I\) , with

    \[ f(x)=f(a)+f'(a)(x-a)+\frac{f''(c)}{2}\,(x-a)^2 . \]

    By hypothesis \(f''(c)\ge 0\) , and \((x-a)^2>0\) . So the last term is nonnegative and

    \[ f(x)\ge f(a)+f'(a)(x-a). \]

    The right side is the tangent line to the graph at \(a\) , so the graph lies on or above every tangent line. Functions with this property are the convex ones; the property is the starting point of the theory developed in Continuous Optimization. If \(f''>0\) throughout, the inequality is strict for \(x\ne a\) .

19. Let \(f(x)=e^{-1/x^2}\) for \(x\ne 0\) and \(f(0)=0\) . Prove that \(f'(0)=0\) . Then show by induction that for \(x\ne 0\) every derivative has the form \(f^{(n)}(x)=q_n(1/x)\,e^{-1/x^2}\) with \(q_n\) a polynomial, and deduce that \(f^{(n)}(0)=0\) for all \(n\) .

    Hint

    The key limit is \(\lim_{x\to 0}x^{-k}e^{-1/x^2}=0\) for every integer \(k\ge 0\) . To prove it put \(u=1/x^2\) and use \(e^u\ge u^m/m!\) for \(u>0\) , which follows from Taylor’s theorem applied to \(e^u\) . The case \(k=1\) gives \(f'(0)=0\) . For the induction, differentiating \(q_n(1/x)e^{-1/x^2}\) produces a polynomial in \(1/x\) times \(e^{-1/x^2}\) ; then \(f^{(n+1)}(0)=\lim_{x\to 0}q_n(1/x)e^{-1/x^2}/x=0\) by the key limit.

20. Let \(f(x)=x^3+x\) . Prove that \(f\) has an inverse \(g:\mathbb R\to\mathbb R\) which is differentiable everywhere, and compute \(g'(0)\) and \(g'(2)\) .

    Solution

    Invertibility. \(f\) is a polynomial, so it is continuous and differentiable, with \(f'(x)=3x^2+1>0\) for all \(x\) . By Corollary 6.8(b) it is strictly increasing, hence one-to-one. Its image is an interval by Corollary 5.8. Since \(f(n)\ge n\) and \(f(-n)\le-n\) for every \(n\in\mathbb N\) , the image is unbounded in both directions, so it is all of \(\mathbb R\) . Thus \(f\) has an inverse \(g:\mathbb R\to\mathbb R\) , continuous by Theorem 5.10.

    Differentiability. Since \(f'(a)\ne 0\) for every \(a\) , Theorem 6.14 applies at every point: \(g\) is differentiable on \(\mathbb R\) and

    \[ g'(f(a))=\frac{1}{3a^2+1}. \]

    Values. \(f(0)=0\) , so \(g(0)=0\) and \(g'(0)=1/(0+1)=1\) . \(f(1)=2\) , so \(g(2)=1\) and \(g'(2)=1/(3+1)=\tfrac14\) .

    As a check, \(g(2.01)\) should be close to \(1+0.01\cdot\tfrac14=1.0025\) ; solving \(x^3+x=2.01\) numerically gives \(x\approx 1.002495\) .

21. Prove that a polynomial of degree \(n\ge 1\) has at most \(n\) distinct real zeros, by induction using Rolle’s theorem. Then outline a proof of L’Hôpital’s rule when \(g(x)\to\infty\) as \(x\to a^+\) : for \(a<x<y\) apply Theorem 6.10 on \([x,y]\) , and write \(f(x)/g(x)\) in terms of the resulting quotient.

    Hint

    A polynomial of degree \(1\) has one zero. If a polynomial \(p\) of degree \(n+1\) had \(n+2\) distinct real zeros, Rolle’s theorem between consecutive zeros would give \(n+1\) distinct zeros of \(p'\) , a polynomial of degree \(n\) . For the second part: for \(a<x<y\) , Theorem 6.10 gives \(c\in(x,y)\) with \(\dfrac{f(x)-f(y)}{g(x)-g(y)}=\dfrac{f'(c)}{g'(c)}\) . Write \(\dfrac{f(x)}{g(x)}=\dfrac{f(x)-f(y)}{g(x)-g(y)}\Bigl(1-\dfrac{g(y)}{g(x)}\Bigr)+\dfrac{f(y)}{g(x)}\) ; first fix \(y\) close to \(a\) so that the quotient of derivatives is within \(\varepsilon\) of \(L\) on \((a,y)\) , then let \(x\to a^+\) so that \(g(y)/g(x)\) and \(f(y)/g(x)\) tend to \(0\) .

22. Let \(f\) be continuous on \([a,b)\) and differentiable on \((a,b)\) , and suppose \(\lim_{x\to a^+}f'(x)=L\) exists. Prove that \(f\) is differentiable at \(a\) with \(f'(a)=L\) . Deduce that a derivative cannot have a removable discontinuity.

    Hint

    For \(x\in(a,b)\) the Mean Value Theorem on \([a,x]\) gives \(c_x\in(a,x)\) with \((f(x)-f(a))/(x-a)=f'(c_x)\) . As \(x\to a^+\) , \(c_x\to a^+\) and \(f'(c_x)\to L\) . For the deduction: if \(f\) is differentiable near \(a\) and \(\lim_{x\to a}f'(x)\) exists, the two one-sided versions give \(f'(a)\) equal to that limit, so \(f'\) is continuous at \(a\) .