## 170- Quiz 10

Quiz 10 is here.

Solutions follow.

Problem 1. Let $f (x) = e^x+x$ . Find $c$ strictly between 0 and 1 with the property that $f'(c)$ equals the slope between the endpoints of $f (x)$ on ${}[ 0, 1]$.
For extra credit, and only if you have finished both problems, use Newton’s method to compute the first 3 significant digits of $c$.

Since $f(x)=e^x+x$, we have that $f(0)=e^0+0=1$ and $f(1)=e^1+1=e+1$. The slope of the line going through $(0,f(0))=(0,1)$  and $(1,f(1))=(1,e+1)$ is

$\displaystyle \frac{(e+1)-1}{1-0}=e$.

The problem is therefore asking us to find a $c$ with $0 and

$f'(c)=e$.

Since $f'(x)=e^x+1$, what we need to do is solve the equation $e^c+1=e$. We have $e^c=e-1$ or

$c=\ln(e-1)$.

If we want to approximate the value of $c$, we use that it is a solution of the equation $e^x+1=e$, that can be written in the form $g(x)=0$, where $g(x)=e^x+1-e$. We use Newton’s method, which says that starting with a guess $x_0$, we can approximate a solution to $g(x)=0$ by improving the guess successively by means of the iteration given by

$\displaystyle x_{n+1}=x_n-\frac{g(x_n)}{g'(x_n)}$, $n=0,1,\dots$

We have $g'(x)=e^x$. Say that $x_0=0$. Then we have:

• $\displaystyle x_{1}=x_0-\frac{g(x_0)}{g'(x_0)}=0-\frac{2-e}{1}=e-2\approx 0.71828183$.
• $\displaystyle x_2=x_1-\frac{g(x_1)}{g'(x_1)}=(e-1)\times e^{-e+2}+e-3\approx 0.55609766$.
• $\displaystyle x_3=x_2-\frac{g(x_2)}{g'(x_2)}\approx 0.54143343$.
• $\displaystyle x_4=x_3-\frac{g(x_3)}{g'(x_3)}\approx 0.54132486$.
• $\displaystyle x_5=x_4-\frac{g(x_4)}{g'(x_4)}\approx 0.54132485$.
• $\displaystyle x_6=x_5-\frac{g(x_5)}{g'(x_5)}\approx 0.54132485$.

Note that if $x=0.54132485$ then we have

$g(x)\approx -7.4505806\times 10^{-9}$,

so this value of $x$ is a fairly decent approximation to $c$.

Problem 2. Find all functions $f(x)$ with the following properties: $f''(x)=\sin(x)+e^x$, $f(0)=1$, $f(1)=0$.

We use that if $g'(x)=h'(x)$ for all $x$, then $g(x)=h(x)+k$ for some constant $k$.

If $f''(x)=\sin(x)+e^x$ then $f'(x)=-\cos(x)+e^x+c$ for some constant $c$. But then $f(x)=-\sin(x)+e^x+cx+d$ for some constant $d$.

To find $c$ and $d$, we use that $f(0)=1$ and $f(1)=0$:

• $1=f(0)=-\sin(0)+e^0+c0+d=1+d$, so $d=0$.
• $0=f(1)=-\sin(1)+e^1+c1+d=-\sin(1)+e+c$, so $c=\sin(1)-e$.

Then the only function satisfying the given conditions is $f(x)=-\sin(x)+e^x+(\sin(1)-e)x$.