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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.

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    $\begingroup$ The instructions call for sketching a pair of graphs. I would actually sketch the line $y=x-2$ and the exponential curve $y=e^{-x/2}$. $\endgroup$ Commented Sep 30, 2020 at 10:47
  • $\begingroup$ The question says "a pair of graphs". For example, $y=x-2$ and $y=e^{-x/2}$. $\endgroup$
    – user65203
    Commented Sep 30, 2020 at 12:35
  • $\begingroup$ I see. Is there any specific reason to prefer sketching $y = x - 2$? $\endgroup$
    – Aidan
    Commented Sep 30, 2020 at 19:17

2 Answers 2

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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.

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  • $\begingroup$ $limx→∞g(x)=2$ 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
    Commented Sep 30, 2020 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 $x \rightarrow \infty$. $\endgroup$
    – Alessio K
    Commented Sep 30, 2020 at 19:27
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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 $2.5$.

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