Dror Bar-Natan: Classes: 2002-03: Math 157 - Analysis I:

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Solution of Term Exam 2

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Problem 1. Prove that there is a real number $x$ so that

$$x^{157}+\frac{157}{1+x^2+\cos^2 x} = 157.$$

If your proof uses the intermediate value theorem, state it clearly and prove that it follows from the postulate P13.

Solution. As a composition/sum/quotient of continuous functions, the left hand side is a continuous function of $x$. The term $\frac{157}{1+x^2+\cos^2 x}$ is bounded by 157 and hence the large $x$ behaviour of the left hand side is dominated by that of $x^{157}$. Thus for large negative $x$ the left hand side goes to $-\infty$ and for large positive $x$ it goes to $+\infty$. Thus by the intermediate value theorem the left hand side must attain the value $157$ for some $x$.

Our proof does use the intermediate value theorem, and hence its statement and proof should be reproduced. See Spivak's chapter 8.

Problem 2.

1. Define in precise terms ``$f$ is differentiable at $a$''.
2. Let
   $$f(x) = \begin{cases} x^2 & x\in{\mathbb{Q}}, \\ 0 & x\not\in{\mathbb{Q}}. \end{cases}$$
   Is $f$ differentiable at 0? If you think it is, prove your assertion and compute $f'(0)$. Otherwise prove that it isn't.

Solution.

1. A function $f$ is said to be differentiable at a point $a$ if the limit
   $$\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$
   exists.
2. According to the definition of differentiability, we consider the limit
   $$\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{h\to 0}\frac{f(h)}{h}.$$
   We claim that this limit is 0 and hence $f'(0)$ exists and is equal to 0. Indeed, Let $\epsilon>0$ be any positive number and set $\delta=\epsilon$. Now if $h$ satisfies $0<\vert h\vert<\delta$ is rational then $\left\vert\frac{f(h)}{h}-0\right\vert=\left\vert\frac{h^2}{h}\right\vert=\vert h\vert<\delta=\epsilon$ and if $h$ satisfies $0<\vert h\vert<\delta$ is irrational then $\left\vert\frac{f(h)}{h}-0\right\vert=\left\vert\frac{0}{h}\right\vert=0<\epsilon$, so in general $0<\vert h\vert<\delta$ implies $\left\vert\frac{f(h)}{h}-0\right\vert<\epsilon$. Thus $\lim_{h\to 0}\frac{f(h)}{h}=0$ as asserted above.

Problem 3. Calculate $dy/dx$ in each of the following cases. Your answer may be in terms of $x$, of $y$, or of both, but reduce it algebraically to a reasonably simple form. You do not need to specify the domain of definition.

$$\vphantom{\frac{a}{b}} x^3 + y^3 = 2 y^4+y^3+xy=1$$

$$\vphantom{\frac{a}{b}} y=x/\sqrt{x^2-4} y=\sin\bigl(\sin(x)\bigr)$$

Solution.

(a)

Differentiating both sides with respect to $x$ we get $3x^2+3y^2y'=0$ and hence $\frac{dy}{dx}=y'=-\frac{x^2}{y^2}$.

(b)

Using the rule for differentiating a quotient, then the chain rule and then simplifying a bit, we get

$$\frac{dy}{dx} = \frac {x'\sqrt{x^2-4}-x\left(\sqrt{x^2-4}\right)'... ...left(x^2-4\right)^{-1/2}\cdot 2x}{x^2-4} =-\frac{4}{\left(x^2-4\right)^{3/2}}.$$

(c)

Differentiating both sides with respect to $x$ we get $4y^3y'+3y^2y'+y+xy'=0$ and hence $$y'=\frac{-y}{x+3y^2+4y^3}$$.

(d)

Using the chain rule, $y'=\cos(\sin(x))\cos(x)$.

Problem 4.

1. Prove that if $f'(x)>0$ on some interval then $f$ is increasing on that interval.
2. Sketch the graph of the function $$f(x)=x+\frac{4}{x^2}$$.

Solution.

1. See Spivak chapter 11.
2. $f(0)$ is not defined; $\lim_{x\to 0}f(x)=+\infty$. The only solution to $f(x)=0$ is $x=-\sqrt[3]{4}$, so the point $(-\sqrt[3]{4}, 0)$ is on the graph. $f'(x)=1-8/x^3$; this is positive when $x>\sqrt[3]{8}=2$ and when $x<0$ and negative when $0<x<2$, so $f$ is increasing when $x>2$ and when $x<0$ and decreasing when $0<x<2$. The derivative is 0 only at $x=2$; right before, the function is decreasing and right after it is increasing. So $x=2$ is a local max and we can compute $f(2)=3$. Finally, $\lim_{x\to\pm\infty}=\pm\infty$ and near $\infty$ our graph $y=f(x)$ is very close to $y=x$, so we arrive at the following graph:

Problem 5. Write a formula for $(f^{-1})''(x)$ in terms of $f'$, $f''$ and $f^{-1}(x)$. Under what conditions does your formula hold?

Solution. From class material we knot that if $f$ is continuous and $1-1$ near $f^{-1}(x)$, differentiable at $f^{-1}(x)$, and $f'(f^{-1}(x))\neq 0$ then $(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$. Using this we get

$$\begin{split}(f^{-1})''(x) &= \left(\frac{1}{f'(f^{-1}(x))}\r... ...)))^2} = -\frac{f''(f^{-1}(x))}{(f'(f^{-1}(x)))^3}. \end{split}$$

In the last chain of equalities we've used the chain rule, for which, in addition to what we already have, we need to know that $f'$ is continuous around $f^{-1}(x)$ and differentiable at $f^{-1}(x)$ and the rule for differentiating a quotient, for which we need nothing new. Hence the full list of conditions needed for aour formula to hold is:

• $f$ is $1-1$ near $f^{-1}(x)$.
• $f$ is differentiable around $f^{-1}(x)$.
• $f'(f^{-1}(x))\neq 0$.
• $f$ is twice differentiable at $f^{-1}(x)$.

an alternative solution:

The results. 86 students took the exam; the average grade is 70.76, the median is 72 and the standard deviation is 18.35.

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