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<c<1 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.

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This entry was posted on Tuesday, November 16th, 2010 at 12:37 am and is filed under 170: Calculus I. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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