# Graph $x = 2 + e^{-\frac{1}{2}x}$

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.

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

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 $$g(3). So sketching both graphs, we can see that there is a solution for $$x\in(0,3)$$. See here.
• $limx→∞g(x)=2$ because $e^\frac{-x}{2}$ has an asymptote at 2, right? With those limits, we are figuring out the end behavior of the function? Just clarifying. Commented Sep 30, 2020 at 19:15
• Yes we are finding the end behavior of the function (although not needed since we have that $g(0)>f(0)$ and $g(3)<f(3)$, so we have solution in the interval (0,3)). $e^{-\frac{x}{2}}$ approaches $0$ for large positive $x$, so it has an asymptote there. Thus $g(x)\rightarrow 2$ as $x \rightarrow \infty$. Commented Sep 30, 2020 at 19:27
For part (iii), you should find that for $$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$$.