Continuity and differentiability of functions

True-False-True-True

In parts a and b, we have \(|f(x)-f(a)|\) lying between \(\pm 39|x-a| .\) As \(x \rightarrow a\) \(\pm 39|x-a| \rightarrow 0\) and hence \(f(x)-f(a) \rightarrow 0,\) so \(f\) is continuous. But it need not be differentiable, for example \(f(x)=|x|\) satisfies \(f(x)-f(y)=|x|-|y| \leq|x-y| \leq 39|x-y| \cdot\) But \(f\) is not differentiable at \(0 .\)

In parts c and d, we have \(\left|\frac{f(x)-f(a)}{x-a}\right| \leq 39|x-a|,\) so by reasoning as for part a, we have \(\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=0,\) i.e., \(f^{\prime}(a)=0\) for all \(a,\) so \(f\) is a constant function.

(i) \(5\). By the intermediate value theorem over the interval \([2,3]\). Also \(f(x)\) may not take the value \(25,\). For a counterexample, take \(f(x)=x^{2}\) for \(x<4\) and \(f(x)=16\) for \(x \geq 4\).

(ii) Both.

(iii) (2,4)

(iv) 4

For parts b, c and d, let \(g(x)=f(x)-x^{2}\). We have \(g(1)=g(2)=g(3)=g(4)=0\). For part b, applying Rolle's theorem to \(g(x)\) gives \(g^{\prime}(x)=0\) for some values of \(x\) in each of the intervals \((1,2),(2,3),(3,4) .\) For these values of \(x,\) we have \(f^{\prime}(x)=g^{\prime}(x)+2 x=2 x\)

For part c, take from part b values \(r \in(2,3)\) and \(s \in(3,4)\) with \(g^{\prime}(r)=0=g^{\prime}(s)\). Applying Rolle's theorem to \(g^{\prime}(x)\) in the interval \([r, s],\) we get for some \(x \in(r, s) \subset(2,4)\) the equality \(g^{\prime \prime}(x)=0\) and so \(f^{\prime \prime}(x)=g^{\prime \prime}(x)+2=2 .\) There need not be an \(x \in(1,2)\) with \(f^{\prime \prime}(x)=2\) i.e., \(g^{\prime \prime}(x)=0 .\) There are many ways to arrange this, for example let \(g(x)=\sin (\pi x) .\) Then

\(g^{\prime \prime}(x)=-\pi^{2} \sin (\pi x),\) which is 0 only when \(x\) is an integer, in particular \(g^{\prime \prime}(x) \neq 0\) for any \(x \in(1,2)\) For part d, note that \(g(x),\) now being a polynomial vanishing at 1,2,3 and \(4,\) must be divisible by \((x-1)(x-2)(x-3)(x-4) .\) So \(g(x),\) if non-zero, must have degree at least 4 Thus \(f(x)=x^{2}\) or a polynomial of degree at least 4

TFFT

i The mean value theorem tells us \(S \subset T\).

ii \(T \subset S\) is false, example \(f(x)=\sin (x) .\) Here \(f^{\prime}(0)=1\) is in \(T\) but not in \(S\).

iii \(T=S=\mathbb{R}\) can happen at points where \(f\) is not differentiable.

iv \(S\) has mean value property, because of continuity. (TODO)

The given function is defined using the two functions \(x^{2}+1\) and \(\tan (x)\).

Both these functions are continuous wherever they are defined. As every irrational number \(z\) has a non terminating, non repeating decimal expansion we see that given any \(\epsilon>0\) there is a rational number \(p\) such that the distance between \(z\) and \(p\) is less than \(\epsilon\).

Using these facts we see that the given function is continuous precisely at those \(x\) in the interval \([0,4 \pi]\) where \(x^{2}+1=\tan (x)\), since \(x^{2}+1\) is positive, it will intersect \(\tan (x)\) exactly once in the intervals \(\left[0, \frac{\pi}{2}\right],\left[\pi, \frac{3 \pi}{2}\right],\left[2 \pi, \frac{5 \pi}{2}\right],\left[3 \pi, \frac{7 \pi}{2}\right],\) as \(\tan (x)\) increases from 0 to \(\infty\) in each of these intervals. \(\tan (x)\) is negative elsewhere in the given domain. So we have 4 points of continuity.

Put \(p(x)=x^2-1\), then \begin{align} \frac{d^{100}}{d x^{100}}\left(\frac{p(x)}{x^{3}-x}\right)=\frac{d^{100}}{d x^{100}}\left(\frac{1}{x}\right) = \frac{-100!}{x^{100}} \end{align} Hence, \(f(x)\) can have degree 0.

(a) \(g^{\prime \prime}(x)=20 x^{3}-120 x^{2}=20 x^{2}(x-6)\) and this changes sign only at \(x=6 .\)

(b) is false. Consider the function above: \(g^{\prime \prime}(0)=0\) but \(g^{\prime \prime}\) does not change sign at \(x=0\).

(c) is true due to continuity. The function \(g\) is continuous since it is differentiable.

Let \(f(x) := \tan^{-1}x \). \begin{align} f^\prime(x) &= \frac{1}{1+x^2} \\ f^{\prime\prime}(x) &= \frac{-2x}{ (1+x^2)^2 } \\ \end{align} In the domain \( (0,\infty) \), the second derivative is strictly negative. This means that \( f(x) \) is strictly concave in this domain. Therefore \( f(x) + f(y) < 2 f( (x+y)/2) \) for any \(x,y \in (0,\infty) \).

(i) Let the width, length and height of the box be \(a,b\) and \(c\), respectively. We are given that \(4(a+b+c)=60\). Volume lies in the interval \( (0,125] \). The volume becomes arbitrarily small as \(a\) tends to zero. To find the maximum volume we use AM-GM inequality:

(ii) The length of the diagonal of the box is given by \( \sqrt{ a^2+b^2+c^2 } \).

Length of diagonal is 13 cm.

(iii) Volume is not uniquely determined.

Option (B). There are unique values of \(a\) and \(b\) for which \(f\) is differentiable at \(x=1\).

For the function to be continuous at \(x=1\), we must have \( a+b = e \). For the function to be differentiable:

\[ e^x = \frac{1}{x} + 2ax + b \;\;\;\text{ at }\; x=1 \]

This means at \( x= 1\), we must have \( a=-1\) and \( b=e+1 \) for the function to be differentiable.

Since the function has odd degree as the highest power, it has at least one real root. To show that the function has exactly one real root we have to show, that it is monotonic in the whole \(\mathbb{R}\). This can be done by showing that \(f'(x)\) is always positive when \(c>1/3\).

\[ f'(x) = 3x^2+2x+c \]

The discriminant of the above quadratic, \(2^2-4\cdot3\cdot c\), is negative when \(c>1/3\). Hence, \(f'(x)\) is always positive in \(\mathbb{R}\).

Since \(f\) is differentiable, it must be continuous. By continuity, there is \(a \in(2,3)\) with \(f(a)=2=f(1)\). Rolle's theorem says that there is \(x \in(1, a)\) with \(f^{\prime}(x)=0\).

If \(f(x)\) is not constant, then it must attain a local maxima or minima at some value \(c\in[a,b]\).

From the condition specified:

\begin{equation} f(c)=f'(c)+f^{\prime\prime}(c) \label{eq:rolle} \tag{1} \end{equation}

If \(f(c)\) is the local maxima, then \(f(c)>0\), \(f'(c)=0\) and \(f^{\prime\prime}(c)<0\). But Eq.\(\eqref{eq:rolle}\) does not hold in that case. A similar argument holds for local minima. Hence, \(f(x)\) is zero in the interval \([a,b]\).

Substitute \(t=\tan^{2} \theta \). Then we have

\begin{align} f'(\tan^2\theta) & = \cos 2\theta(1+\tan \theta \tan 2\theta ) \\ f'(\tan^2\theta) & = \frac{1-\tan^2\theta}{1+\tan^2\theta} \left( 1 + \frac{2\tan^2\theta}{1-\tan^2\theta} \right) \\ f'(t) & = \frac{1-t}{1+t} \left( 1 + \frac{2t}{1-t} \right) \quad\text{ where } t=\tan^2 \theta \\ f'(t) & = 1 \\ f(t) & = t + constant \end{align}

Hence the answer is \(1-\cos \frac{\pi}{12}\).

15,6

This problem is similar to the last one. We need to show that the function is monotonic in some interval.

Consider the function: \begin{align} f(x)&= e^x -\frac{x}{2017} - 1 \\ f'(x)&= e^x -\frac{1}{2017} \end{align}

The derivative is positive when \(x> x_0=-\log 2017\).

• We have \(f(x_0)< 0\) and \( f(-\infty)>0 \) so there is one solution in the interval \((-\infty,x_0)\).
• \( x = 0 \) is another solution.

Hence, there are two solutions to the equation.

(i) \(n\) at \(x=1,2, \ldots, n\).

(ii) \(n>4,\) because \(\lim_{x \rightarrow \pm \infty} f(x)=0\) and for this to happen, the degree of the denominator of \(f(x)\) must be greater than that of the numerator.

(iii) \(x=n-1,\) because \(f(x)\) is positive for \(x> n\) and \(f(x)\) changes sign precisely when it passes through \(x=1,2 \ldots, n .\) Note that the sign of \(f(x)\) for \(x<0\) and for \(x \in(0,1)\) depends on the parity of \(n\)

(iv) More than \(1 .\) Note that \(f(x)=0\) only at \(x=0,\) with multiplicity \(4 .\) Without loss of generality, let \(n\) be even. Now \(f(x)> 0\) for \(x<1\) except at \(x=0\) and \(f\) has all derivatives for \(x<1\). Due to the multiple root at \(x=0,\) the graph of \(f\) must be concave up, that is \(f^{\prime \prime}(x)> 0\) ) near \(x=0\). Further, as \(x \rightarrow-\infty,\) the values of \(f(x)\) stay positive and \(\rightarrow 0\).

Therefore, as one traces the graph from from the origin to the left, it must become concave down at least once and eventually concave up again so as to approach the X-axis from above.

We have \(f(0)=1, f(1)=a+\frac{1}{2}\) and \(f^{\prime}(x)=a-\frac{1}{(x+1)^{2}}\) Over the interval \([0,1],\) the value of \(f^{\prime}(x)\) increases from \(a-1\) at \(x=0\) to \(a-\frac{1}{4}\) at \(x=1\)o

What happens to the sign of \(f^{\prime}(x)\)? We do a case by case analysis.

Case 1. Suppose \(a \leq 1 / 4 .\) Because \(1 /(x+1)^{2} \geq 1 / 4\) on the interval \([0,1], f^{\prime}(x) \leq 0,\) so the maximum is at 0 and the minimum is at \(x=1\). So the difference is \(1-(1 / 2+a)=\) \(1 / 2-a \geq 1 / 4 \geq 1 / 12\)

Case 2. Suppose \(a \geq 1\). Then \(f^{\prime}(x) \geq 0\) on the interval [0,1] , so maximum is at 1 and minimum at 0. We get \(a+1 / 2-1=a-1 / 2 \geq 1 / 2 \geq 1 / 12\)

Case 3. Suppose \(1 / 4 \leq a \leq 1 .\) Now \(f^{\prime}(x)=0\) at \(\tilde{x}=\frac{1}{\sqrt{a}}-1 .\) For this range of \(a, \tilde{x} \in[0,1]\) In the interval \([0, \tilde{x}], f^{\prime}(x) \leq 0\) and in the interval \([\tilde{x}, 1], f^{\prime}(x) \geq 0 .\) Now we make two sub-cases depending on the endpoint at which the maximum occurs.

Case 3a. Suppose \(1 / 4 \leq a \leq 1 / 2\). Then \(f(0) \geq f(1)\). So minimum is at \(\tilde{x},\) maximum is at \(x=0 . f(\tilde{x})=\sqrt{a}-a+\sqrt{a}=2 \sqrt{a}-a\). So the difference between maximum and minimum is \(1+a-2 \sqrt{a}=(1-\sqrt{a})^{2}\). This is smallest when \(\sqrt{a}\) is closest to 1 and so \((1-\sqrt{a})^{2} \geq(1-1 / \sqrt{2})^{2}=3 / 2-\sqrt{2} .\) This is bigger than \(1 / 12\) since \(\left(\frac{3}{2}-\frac{1}{12}\right)=17 / 12\) and \(17^{2}=289 \geq 2 \times 12^{2}\)

Case 3b. Suppose \(1 / 2 \leq a \leq 1\). Now \(f(1) \geq f(0)\). Max is at 1 and minimum is at \(\tilde{x}\). The difference is \(a+1 / 2-\sqrt{a}+a-\sqrt{a}=2 a-2 \sqrt{a}+1 / 2=\left(\sqrt{2 a}-\frac{1}{\sqrt{2}}\right)^{2} .\) By a calculation similar to the above it is bigger than \(1 / 12\).

The only problem is at \(x=0 .\) For continuity, \(\ln (0+1)=C\).

For \(f^{\prime}(0)\) to exist, \(f\) must be continuous and the left and right derivatives of \(f\) at \(x=0\) (which are easily seen to exist) must match, that is \(5=B\).

For \(f^{\prime \prime}(0)\) to exist, \(f^{\prime}(0)\) must exist and left and right derivatives of \(f^{\prime}\) at \(x=0\) must match, i.e. \(2 A=-5^{2} .\) So \(A=-\frac{25}{2}, B=5, C=0\).