Exercises: Derivatives 3 – "Math for Non-Geeks"

Rolle's theorem and the intermediate value theorem

Exercise

Is there a $c\in ℝ$ such that $f(x)=x^3-3x+c$ has two distinct zeros in $[-1,1]$?

Solution

Suppose there is a $c\in ℝ$ such that $f$ has two zeros $x_0,x_1\in [-1,1]$. Since $f$ is continuous on $[x_0,x_1]\subseteq [-1,1]$ and differentiable on $(x_0,x_1)\subseteq (-1,1)$, according to Rolle's theorem there is a $\xi \in (x_0,x_1)$ with $f^{\prime }(\xi )=0$. But now there is for every $c\in ℝ$: $f^{\prime }(x)=3x^2-3=0\iff x^2=1\iff x=\pm 1\notin (x_0,x_1)$. So $f$ cannot have two zeros.

Exercise

Let $f:ℝ\to ℝ$ be differentiable and $f^{\prime }$ continuous. Let further $f(0)=0$, $f(1)=2$ and $f(2)=1$. Show that $f^{\prime }$ has at least one zero.

Proof

$f$ is continuous on $[0,1]\subset ℝ$ and $f(0)=0<1<2=f(1)$. Therefore, according to the intermediate value theorem, there is an $x_0\in [0,1]$ with $f(x_0)=1$.

Further, $f$ is continuous on $[x_0,2]\subset ℝ$, differentiable on $(x_0,2)\subset ℝ$ and $f(x_0)=1=f(2)$. According to Rolle's theorem, there is hence a $\xi \in (x_0,2)$ with $f^{\prime }(\xi )=0$. So $f$ has at least one zero (on $ℝ$).

Mean value theorem

Exercise (An easy application of the mean value theorem)

Let $f:[0,1]\to ℝ$, $f(x)=2\ln (x^3+\sqrt{2-x^2})$. Show that there is a $y\in (0,1)$ with $f^{\prime }(y)=\ln (2)$?

Solution (An easy application of the mean value theorem)

$f$ is continuous on $[0,1]$ and differentiable on $(0,1)$, as a composition of continuous and differentiable functions, respectively. Thus, the mean value theorem is applicable. There is hence a $y\in (0,1)$ with

$f^{\prime }(y)=\frac{f(1)-f(0)}{1-0}=2\ln (2)-\ln (2)=\ln (2)$

Exercise (Useful inequality 1)

Show that: For all $x\in ℝ$ there is

$1+x\le e^x\le x{e}^x+1$

Solution (Useful inequality 1)

Proof step: $1+x\le e^x$

Case 1: $x>0$

We define $f:[0,x]\to ℝ$ by $f(t)=e^t$. Then $f$ is continuous, and on $(0,x)$ differentiable. Thus, the mean value theorem is applicable, and there exists a $\xi \in (0,x)$ with

$f^{\prime }(\xi )=\frac{f(x)-f(0)}{x-0}⟺e^{\xi }=\frac{e^x-e^0}{x-0}=\frac{e^x-1}{x}$

Now, because of $\xi \in (0,x)$,

$\frac{e^x-1}{x}=e^{\xi }\stackrel{\xi >0}{>}{e}^0=1⟺e^x-1>x⟺e^x>x+1$

Case 2: $x=0$

Here we have $1+0=e^0$, i.e., equality.

Case 3: $x<0$

Again, by the mean value theorem there is a $\xi \in (x,0)$ with

$\frac{e^x-1}{x}=e^{\xi }\stackrel{\xi <0}{<}{e}^0=1\stackrel{x<0}{⟺}{e}^x-1>x⟺e^x>x+1$

So there is $1+x\le e^x$ for all $x\in ℝ$

Proof step: $e^x\le x{e}^x+1$

Here we show only the case $x>0$:

We again define $f:[0,x]\to ℝ,f(t)=e^t$. Then $f$ is continuous, and on $(0,x)$ differentiable. According to the mean value theorem, there exists a $\xi \in (0,x)$ with

$\frac{e^x-1}{x}=e^{\xi }\stackrel{\xi <x}{<}{e}^x⟺e^x-1<x{e}^x⟺e^x<x{e}^x+1$

For $x=0$ there is again equality, and for $x<0$ the statement follows analogously with the mean value theorem.

Hint

From the first inequality, the transition to $-x$ for $x<1$ still allows for the inequality:

$e^x\le \frac{1}{1-x}$

Exercise (Useful inequality 2)

Show that: For $0<x<\frac{\pi }{2}$ there is

$\sin (x)<x<\tan (x)$

Solution (Useful inequality 2)

Proof step: $\sin (x)<x$

Let $x\in (0,\frac{\pi }{2})$. Then the sine function on $[0,\frac{\pi }{2}]$ is continuous and on $(0,\frac{\pi }{2})$ it is differentiable. With the mean value theorem there is a $\xi \in (0,\frac{\pi }{2})$ with

$f^{\prime }(\xi )={\sin }^{\prime }(\xi )=\frac{\sin (x)-\sin (0)}{x-0}=\frac{\sin (x)}{x}$

But now there is $f^{\prime }(\xi )=\cos (\xi )<1$ for $\xi \in (0,\frac{\pi }{2})$. Thus, we have

$\frac{\sin (x)}{x}<1⟺\sin (x)<x$

Proof step: $x<\tan (x)$

Let $x\in (0,\frac{\pi }{2})$. Then the tangent function on $[0,\frac{\pi }{2}]$ is continuous and on $(0,\frac{\pi }{2})$ it is differentiable. With the mean value theorem there is a $\xi \in (0,\frac{\pi }{2})$ with

$f^{\prime }(\xi )=1+{\tan }^2(\xi )=\frac{\tan (x)-\tan (0)}{x-0}=\frac{\tan (x)}{x}$

But now there is $f^{\prime }(\xi )=1+{\tan }^2(\xi )>1$ for $\xi \in (0,\frac{\pi }{2})$. Thus, we have

$\frac{\tan (x)}{x}>1⟺\tan (x)>x⟺x<\tan (x)$

Hint

The inequality can be further extended to all $x\in ℝ$:

$|\sin (x)|\le |x|\le |\tan (x)|$

Where equality only holds at $x=0$.

Exercise (Implication of the mean value theorem)

Show by the mean value theorem that:

Let $f,g:[a,b]\to ℝ$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Furthermore, let $f(a)\ge g(a)$ and $f^{\prime }\ge g^{\prime }$ hold on $(a,b)$. Then, there is $f\ge g$ on $[a,b]$.

As an application: prove the following generalization of the Bernoulli inequality: For $\alpha >1$ and all $x>0$ there is $(1+x{)}^{\alpha }\ge 1+\alpha x$.

How to get to the proof? (Implication of the mean value theorem)

We present three different possible solutions: One with use of the mean value theorem, one via the monotonicity criterion and one via the fundamental theorem of calculus. Within all three, we make use of the auxiliary function $h=f-g$.

Proof (Implication of the mean value theorem)

We consider the auxiliary function

$h:[a,b]\to ℝ,h(x):=f(x)-g(x)$

This function is continuous and differentiable on $(a,b)$. Further there is

1. $h(a)=f(a)-g(a)\ge 0$
2. $h^{\prime }(x)=f^{\prime }(x)-g^{\prime }(x)\ge 0$ for all $x\in (a,b)$

1st way: By the mean value theorem

By the mean value theorem, for all $x\in (a,b]$ there is a $\xi \in (a,x)$ with

$\frac{h(x)-h(a)}{x-a}=h^{\prime }(\xi )\stackrel{2.}{\ge }0\Rightarrow h(x)-h(a)\ge 0\Rightarrow h(x)\ge h(a)\stackrel{1.}{\ge }0\Rightarrow f(x)\ge g(x)$ for all $x\in [a,b]$

2nd way: By the monotonicity criterion

It follows from 1 that $h$ is monotonically increasing on $(a,b)$ (even on $[a,b]$).

$\Rightarrow h(x)\ge h(a)\stackrel{2.}{\ge }0$ for all $x\in [a,b]\iff f(x)\ge g(x)$ for all $x\in [a,b]$

3rd way: By the fundamental theorem of calculus

By assumption, $h$ is integrable and because of the monotonicity of the integral there is for all $x\in [a,b]$:

${\int }_a^x{h}^{\prime }(t)\mathrm{d}t\stackrel{2.}{\ge }{\int }_a^{x}0\mathrm{d}t=0.$

But now, by the fundamental theorem of calculus

${\int }_a^x{h}^{\prime }(t)\mathrm{d}t=h(x)-h(a)\ge 0\Rightarrow h(x)\ge h(a)\stackrel{1.}{\ge }0\Rightarrow f(x)\ge g(x)\mathrm{\forall }x\in [a,b]$

Concerning the application exercise: We define

$f:[0,\mathrm{\infty })\to ℝ,f(x)=(1+x{)}^{\alpha }=\exp (\alpha \ln (1+x))$

and

$g:[0,\mathrm{\infty })\to ℝ,f(x)=1+\alpha x$

Then $f$ and $g$ are continuous on $[0,\mathrm{\infty })$ and differentiable on $(0,\mathrm{\infty })$ with

$f^{\prime }(x)=\exp (\alpha \ln (1+x))\cdot \frac{\alpha }{1+x}=(1+x{)}^{\alpha }\cdot \frac{\alpha }{1+x}=\alpha (1+x{)}^{\alpha -1}$

and

$g^{\prime }(x)=\alpha$

Furthermore, there is $f(0)=0^{\alpha }=\exp (0)=1\ge 1+\alpha \cdot 0$. Since the exponential function is strictly monotonically increasing, there is for all $x\in (0,\mathrm{\infty })$:

$f^{\prime }(x)=(1+x{)}^{\alpha -1}=\alpha \underset{\ge \exp (0)=1}{\underbrace{\exp ((\alpha -1)\underset{\ge \ln (1)=0}{\underbrace{\ln (1+x)}})}}\ge \alpha \cdot 1=\alpha =g^{\prime }(x)$

With the proven statement we hence get for all $x\in [0,\mathrm{\infty })$:

$f(x)\ge g(x)⟺(1+x{)}^{\alpha }\ge 1+\alpha x$

Hint

If we even have $f(a)>g(a)$ and $f^{\prime }>g^{\prime }$ on $(a,b)$, then we have $f>g$ on $[a,b]$.

Hint

The generalized Bernoulli inequality can even be shown for all $x>-1$. Equality only holds in the case $x=0$.

Exercise (mean valuer theorem for continuously differentiable functions)

Let $f:[a,b]\to ℝ$ be continuous, and continuously differentiable on $(a,b)$. Show that there is a $\xi \in (a,b)$ with $f^{\prime }(\xi )=\frac{f(b)-f(a)}{b-a}$, using the intermediate value theorem.

Solution (mean valuer theorem for continuously differentiable functions)

We consider any function $f$ with the given properties and the secant through the points $(a,f(a))$ and $(b,f(b))$. The slope of the secant is given by the difference quotient $\frac{f(b)-f(a)}{b-a}$. Next we look at the slope of the graph, i.e. the derivative values of the function on the interval $(a,b)$.

Case 1: The function graph is a straight line.

Then the derivative function is constant and consequently there is $\frac{f(b)-f(a)}{b-a}=f^{\prime }(\xi )$ for all $\xi \in (a,b)$.

Case 2: The function graph is not a straight line.

Then a $\lambda \in (a,b)$ must exist with $\frac{f(b)-f(a)}{b-a}>f^{\prime }(\lambda )$ or $\frac{f(b)-f(a)}{b-a}<f^{\prime }(\lambda )$ so that $f$ has no straight line as a function graph. It follows in turn that a $\mu \in (a,b)$ exists with $\frac{f(b)-f(a)}{b-a}<f^{\prime }(\mu )$ or $\frac{f(b)-f(a)}{b-a}>f^{\prime }(\mu )$, because otherwise the graph at $b$ can never take the function value $f(b)$. So in total there exist $\lambda ,\mu \in (a,b)$ with $f^{\prime }(\lambda )<\frac{f(b)-f(a)}{b-a}<f^{\prime }(\mu )$. According to the intermediate value theorem, which is applicable here to the derivative function because it is continuous, there is now a $\xi \in (a,b)$ with $f^{\prime }(\xi )=\frac{f(b)-f(a)}{b-a}$.

So in every m case there is a $\xi \in (a,b)$ with $f^{\prime }(\xi )=\frac{f(b)-f(a)}{b-a}$.

Exercise (Unbounded derivative and uniform continuity)

Let $f:ℝ\to ℝ$ be a differentiable function with $\underset{x\to \mathrm{\infty }}{lim}{f}^{\prime }(x)=\mathrm{\infty }$. Prove that then $f$ is not a uniformly continuous function.

Proof (Unbounded derivative and uniform continuity)

Let $f:ℝ\to ℝ$ be a differentiable function with $\underset{x\to \mathrm{\infty }}{lim}{f}^{\prime }(x)=\mathrm{\infty }$. To prove that $f$ is not uniformly continuous, it must be shown that there is an $ϵ>0$, such that for all $\delta >0$ there are real numbers $x,y\in ℝ$ with $|x-y|<\delta$ and $|f(x)-f(y)|\ge ϵ$.

Choose $ϵ=\frac{1}{2}$. Now let $\delta >0$ be arbitrary. Because of $\underset{x\to \mathrm{\infty }}{lim}{f}^{\prime }(x)=\mathrm{\infty }$ there is an $x_0\in ℝ$ with $f^{\prime }(x)\ge \frac{1}{\delta }$ for all $x\in ℝ$ with $x\ge x_0$. According to the mean value theorem equation there is then for all $x\in ℝ$ with $x\ge x_0$:

$\begin{array}{cc}& f(x)-f(x_0)\ge \frac{1}{\delta }\cdot (x-x_0)\\ \Rightarrow & f(x)\ge f(x_0)+\frac{1}{\delta }\cdot (x-x_0)\ge f(x_0)\end{array}$

Now choose $x:=x_0+\frac{\delta }{2}>x_0$. There is by the above estimation $f(x)=f(x_0+\frac{\delta }{2})\ge f(x_0)$ and we get:

$|f(x_0+\frac{\delta }{2})-f(x_0)|=f(x_0+\frac{\delta }{2})-f(x_0)\ge \frac{1}{\delta }\cdot ((x_0+\frac{2}{\delta })-x_0)=\frac{1}{\delta }\cdot \frac{\delta }{2}=\frac{1}{2}=ϵ$

Thus $f$ is not uniformly continuous.

Exercise (Application of the second mean value theorem)

Let $f,g:(a,b)\to ℝ$ be differentiable. Further let $|f^{\prime }(x)|\ge |g^{\prime }(x)|$ for all $x\in (a,b)$. Show that then also

$|f(y)-f(x)|\ge |g(y)-g(x)|$

holds for all $x,y\in (a,b)$.

Proof (Application of the second mean value theorem)

Let $x,y\in (a,b)$ be arbitrary with $x<y$. Then $f$ and $g$ are continuous on $[x,y]$ and differentiable on $(x,y)$ according to the assumption. Then, with the second mean value theorem, there is a $\xi \in (x,y)$ with

$\frac{f(y)-f(x)}{g(y)-g(x)}=\frac{f^{\prime }(\xi )}{g^{\prime }(\xi )}⟹\left|\frac{f(y)-f(x)}{g(y)-g(x)}\right|=\left|\frac{f^{\prime }(\xi )}{g^{\prime }(\xi )}\right|$

Since by assumption $|f^{\prime }(x)|\ge |g^{\prime }(x)|$ holds for all $x\in (a,b)$ , we get $\left|\frac{f^{\prime }(\xi )}{g^{\prime }(\xi )}\right|\ge 1$.

From this we obtain

$\left|\frac{f(y)-f(x)}{g(y)-g(x)}\right|\ge 1⟺|f(y)-f(x)|\ge |g(y)-g(x)|$

Lipschitz-continuity of functions

Exercise (Lipschitz-continuity of functions)

Show by the mean value theorem (using the implication function about Lipschitz continuous functions), that the following functions are Lipschitz continuous. Determine in addition some suitable Lipschitz constants.

1. $f:ℝ\to ℝ,f(x)=\arctan (x)$
2. $g:ℝ\to ℝ,g(x)=\tanh (x)$
3. $h:ℝ\to ℝ,h(x)=\sin (2x)+\cos (3x)$

Solution (Lipschitz-continuity of functions)

Part 1: For all $x\in ℝ$ there is

$|f^{\prime }(x)|=\frac{1}{1+x^2}\stackrel{x^2\ge 0}{\le }\frac{1}{1+0}=1$

So $f$ has a bounded derivative, and is therefore Lipschitz continuous. Further for all $x,y\in ℝ$ there is

$|f(x)-f(y)|\le \underset{=1}{\underbrace{\underset{x\in ℝ}{sup}|f^{\prime }(x)|}}\cdot |x-y|$

Therefore, $L_1=1$ is an appropriate Lipschitz constant.

Part 2: Here, for all $x\in ℝ$ there is:

$|g^{\prime }(x)|=\frac{1}{{\cosh }^2(x)}\stackrel{\cosh \ge 1}{\le }\frac{1}{1}=1$

Thus, by boundedness of derivative, $g$ is Lipschitz-continuous, as well. Moreover, there is for all $x,y\in ℝ$

$|g(x)-g(y)|\le \underset{=1}{\underbrace{\underset{x\in ℝ}{sup}|g^{\prime }(x)|}}\cdot |x-y|$

Hence, $L_2=1$ is an appropriate Lipschitz constant here.

Part 3: finally there is for all $x\in ℝ$:

$|h^{\prime }(x)|=|2\cos (2x)-3\sin (3x)|\underset{\text{inequality}}{\stackrel{\text{triangle}}{\le }}2|\cos (2x)|+3|\sin (3x)|\stackrel{\sin ,\cos \le 1}{\le }2+3=5$

So also $h$ is Lipschitz-continuous, and for all $x,y\in ℝ$ there is

$|h(x)-h(y)|\le \underset{=5}{\underbrace{\underset{x\in ℝ}{sup}|h^{\prime }(x)|}}\cdot |x-y|$

Hence, $L_3=5$ is an appropriate Lipschitz constant.

Exercise (Lipschitz-continuity of functions)

Let $f:[a,b]\to ℝ$ be a continuous function that is differentiable on $(a,b)$. Let $m,M\in ℝ$ be any two real numbers such that $m\le f^{\prime }(x)\le M$ for all $x\in (a,b)$. Prove that for all $x,y\in (a,b)$ the following estimate holds:

$m\cdot (x-y)\le f(x)-f(y)\le M\cdot (x-y)$

Proof (Lipschitz-continuity of functions)

Let $f:[a,b]\to ℝ$ be a continuous function with the properties of the problem. Let $y,x\in (a,b)$ be arbitrary. If $y=x$ then $x-y=0$ and $f(x)-f(y)=0$, so the above estimate is satisfied. Let therefore in the following $x\ne y$.

Proof step: $m\cdot (x-y)\le f(x)-f(y)$

Proof by contradiction: Let $m\cdot (x-y)>f(x)-f(y)$. Thus $m>\frac{f(x)-f(y)}{x-y}$. It follows from the mean value theorem that there is an $\tilde{x}\in [min\{x,y\},max\{x,y\}]$ with $f^{\prime }(\tilde{x})=\frac{f(x)-f(y)}{x-y}$. Thus $m>\frac{f(x)-f(y)}{x-y}=f^{\prime }(\tilde{x})$ is inconsistent with $m\le f^{\prime }(x_0)$ for all $x_0\in (a,b)$.

Proof step: $f(x)-f(y)\le M\cdot (x-y)$

Proof by contradiction: Let $f(x)-f(y)>M\cdot (x-y)$. Thus $\frac{f(x)-f(y)}{x-y}>M$. It follows from the mean value theorem that there is an $\tilde{x}\in [min\{x,y\},max\{x,y\}]$ with $f^{\prime }(\tilde{x})=\frac{f(x)-f(y)}{x-y}$. Thus $f^{\prime }(\tilde{x})=\frac{f(x)-f(y)}{x-y}>M$ is inconsistent with $f^{\prime }(x_0)\le M$ for all $x_0\in (a,b)$.

Exercise (Local Lipschitz continuity)

Let $f:D\to ℝ$ be a continuously differentiable function, where $D$ is an open subset of $ℝ$. Prove that $f$ is locally Lipschitz continuous. That means, you need to prove that for all $x_0\in D$ there exists an $ϵ>0$ such that $f$ is Lipschitz continuous on $(x_0-ϵ,x_0+ϵ)\cap D$.

Proof (Local Lipschitz continuity)

Let $f:D\to ℝ$ be a continuously differentiable function, where $D\subseteq ℝ$ is open. Let $x_0\in ℝ$ be arbitrary. Since $D$ is open, there is a $\delta >0$ such that $(x_0-\delta ,x_0+\delta )\subseteq D$. Choose $ϵ:=\frac{1}{2}\cdot \delta$. Then $[x_0-ϵ,x_0+ϵ]\subseteq D$. Since $f^{\prime }$ is continuous, $f^{\prime }$ is a bounded function on $[x_0-ϵ,x_0+ϵ]$.

Now we have already proved that differentiable functions with bounded derivatives are Lipschitz continuous. Thus $f$ is Lipschitz continuous on $[x_0-ϵ,x_0+ϵ]$ and thus also on $(x_0-ϵ,x_0+ϵ)\cap D$.