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

Question

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.

Answer 1

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.

Answer 2

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 $2.5$.