Dror Bar-Natan: Classes: 2002-03: Math 157 - Analysis I: (136) Next: Integration
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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

$\displaystyle 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

    $\displaystyle 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.


  1. A function $ f$ is said to be differentiable at a point $ a$ if the limit

    $\displaystyle \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} $

  2. According to the definition of differentiability, we consider the limit

    $\displaystyle \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.

$\displaystyle \vphantom{\frac{a}{b}}$   (a)$\displaystyle  $ $\displaystyle x^3 + y^3 = 2$ (c)$\displaystyle  $ $\displaystyle y^4+y^3+xy=1$    
$\displaystyle \vphantom{\frac{a}{b}}$   (b)$\displaystyle  $ $\displaystyle y=x/\sqrt{x^2-4}$ (d)$\displaystyle  $ $\displaystyle y=\sin\bigl(\sin(x)\bigr)$    


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}$.

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

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

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

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 $ \displaystyle


  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{displaymath}\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}\end{displaymath}    

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:

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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Dror Bar-Natan 2002-12-09