I'm not sure how to begin/approach/do this problem or apply the theorem.
You must deduce an equation for f(x) from the given graph. It is the graph of sin(x), shifted 1 place along positive x, and shifted 2 placed along positive y, and squished/expanded by a factor of... something! You must then inject that equation into the integral:
$\displaystyle g(x) = \int_{-3}^{2x-1}f(t)dt $
And carry out the integration to get a function in x for g(x).
The second fundemental theorem is
$\displaystyle \frac{d}{dx} \int_a^x f(t)\, dt = f(x)$
or if
$\displaystyle g(x) = \int_a^x f(t)\, dt $ then $\displaystyle g'(x) = f(x)$
However to use this theorem, it must look exactly of this form. What you have is
$\displaystyle g(x) = \int_{-3}^{2x-1} f(t)\, dt$
so use the chain rule. Let $\displaystyle u = 2x-1$
$\displaystyle \frac{d}{dx} \int_{-3}^{u} f(t)\, dt = \frac{d}{du} \int_{-3}^{u} f(t)\, dt \cdot \frac{d u}{dx}$
Can you continue?
ok i think i may have got it...
$\displaystyle f'(x)=g(x)$
so by the fundamental theorem
$\displaystyle f'(x)= f(u) du/dx $
$\displaystyle f'(x)= g(2x-1)* 2$
which would mean
$\displaystyle g(x)= (1/2) \int_{-3}^u u du $
u=2x-1
so when x=1/2, u=-1; and when x=-3, u=-7
changing the limits would change the domain to [-1, -7]?!
If $\displaystyle g(x) = \int_{-3}^{2x-1} f(t) \, dt$ then $\displaystyle g' (x)= 2f(2x-1)$, not $\displaystyle f'(x)=g(x)$!
Now we will use $\displaystyle g' (x)= 2f(2x-1)$, to get an idea of the domain of $\displaystyle g(x)$, i.e. places where $\displaystyle g$ is defined. From your graph $\displaystyle f(t)$ is only defined on
$\displaystyle -3 \le t \le 5$ so $\displaystyle f(2x-1)$ is only defined on $\displaystyle -3 \le 2x-1 \le 5\;\; \Rightarrow\;\; -1 \le x \le 3$
So in this interval $\displaystyle f(2x-1)$ is defined so $\displaystyle g'(x)$ is defined and since
$\displaystyle g(x) = \int_{-3}^{2x-1} f(t)\,dt$ then so is $\displaystyle g(x)$.
The maximum (or minimum) will occur at a critical point, a place where $\displaystyle g' = 0 \; \text{or}\; g' DNE$. So
$\displaystyle g'(x) = 2 f(2x-1)$ and so $\displaystyle g'(x) = 0 \; \text{when}\;2 f(2x-1) = 0$ and from the graph $\displaystyle f(0)=0$ which means that $\displaystyle x = \frac{1}{2}$