1
$\begingroup$

My question is specific to part 2.

The constant $a$ is such that $$\int_0^a x\mathrm e^{\frac12x}\,\mathrm dx=6\text.$$

  1. Show that $a$ satisfies the equation $$x = 2 + \mathrm e^{-\frac{1}{2}x}\text.$$
  2. By sketching a suitable pair of graphs, show that this equation has only one root.
  3. Verify by calculation that this root lies between $2$ and $2.5$.
  4. Use an iterative formula based on the equation in part 1 to calculate the value of $a$ correct to $2$ decimal places. Give the result of each iteration to $4$ decimal places.

I have successfully changed the integral into the form given: $$x = 2 + \mathrm e^{-\frac{1}{2}x}$$

Now for the sketching. I am very confused about how I'm supposed to sketch this graph. I know the general graph of the exponential function, but in $x = 2 + \mathrm e^{-\frac{1}{2}x}$, I don't have a $y$!

Can someone go into the details? I can't wrap my head around most of this just yet, so the details would help me see light.

$\endgroup$
3
  • 1
    $\begingroup$ The instructions call for sketching a pair of graphs. I would actually sketch the line $y=x-2$ and the exponential curve $y=e^{-x/2}$. $\endgroup$ – Barry Cipra Sep 30 '20 at 10:47
  • $\begingroup$ The question says "a pair of graphs". For example, $y=x-2$ and $y=e^{-x/2}$. $\endgroup$ – Yves Daoust Sep 30 '20 at 12:35
  • $\begingroup$ I see. Is there any specific reason to prefer sketching $y = x - 2$? $\endgroup$ – Aidan Sep 30 '20 at 19:17
1
$\begingroup$

You can sketch $f(x)=x$. Clearly $f(0)=0$ and $f(3)=3$.

Now lets look at $g(x)=2+e^{-\frac{x}{2}}$. We have $g(0)=3$, $g(3)=2+e^{-\frac{3}{2}}<3,$$\lim_{x\rightarrow\infty}g(x)=2$ and $\lim_{x\rightarrow-\infty}g(x)=\infty.$

Then $g(0)>f(0)$ and $g(3)<f(3)$. So sketching both graphs, we can see that there is a solution for $x\in(0,3)$. See here.

$\endgroup$
2
  • $\begingroup$ $limx→∞g(x)=2$ because $e^\frac{-x}{2}$ has an asymptote at 2, right? With those limits, we are figuring out the end behavior of the function? Just clarifying. $\endgroup$ – Aidan Sep 30 '20 at 19:15
  • $\begingroup$ Yes we are finding the end behavior of the function (although not needed since we have that $g(0)>f(0)$ and $g(3)<f(3)$, so we have solution in the interval (0,3)). $e^{-\frac{x}{2}}$ approaches $0$ for large positive $x$, so it has an asymptote there. Thus $g(x)\rightarrow 2$ as $x \rightarrow \infty$. $\endgroup$ – Äres Sep 30 '20 at 19:27
1
$\begingroup$

For part (iii), you should find that for $g(x) = x - \left(2 + e^{-\frac{1}{2}x} \right)$, $g(2) > 0$ and $g(2.5) < 0$. Then use IVT to show the existence of a root between $2$ and $2.5$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.