Solving a maximum minimum problem 
This is a problem I did in one of my test in high school, I completely forgot how to do this, there were two questions stated a) and b).
My attempt will be to try and find the function of the graph and integrate, then I would try to differentiate it and set it to 0, and try to find that value of of x? I am not quite sure how to fo part a), I think I can try to do part b) which is just integrate on that interval of the function. 
Could anyone help me? Thank you
 A: You can set $g'(x)=\dfrac{d}{dx}\int_{-4}^xf(t)dt$, which is $\dfrac{d}{dx}(F(x)-F(-4))$, if we let $F$ be an antiderivative of $f$. Then the derivative is clearly just $f(x)$.
So, where does $g'(x)=f(x)=0$? According to the graph, this occurs for $x=1$.
Note that part (b) is asking for a function that gives the area bounded by $f$ and the $x$-axis from $-4$ to $x$. You can easily plot points to see that, for example, $(0,-4)$ is on the graph.
A: For part (a), we have that $g'(x) = f(x)$. 
$g(x)$ has critical points whenever $g'(x)$= $f(x)$ =$0$. From the graph, this would correspond with $x=1$. 
Since $g''(x)$=$f'(x)$, and $f'(1)$>$0$ (Why?), we know that $g''(1)$>$0$, so we know that $g(x)$ has a local minimum at $x=1$.
To test whether this is indeed the global minimum of $g(x)$, we must check the value of $g(x)$ at $x=1$ and also check the value of $g(x)$ at the endpoints of the interval.
$g(1)$=$\int_{-4}^{1}$ $f(t)$$dt$=$-4.5$
$g(-4)$=$\int_{-4}^{-4}$ $f(t)$$dt$=$0$
$g(4)$=$\int_{-4}^{4}$ $f(t)$$dt$=$0$
(Check these for yourself!).
So it is clear that $g(x)$ has a global minimum at $x=1$.
For part (b), there is no clear-cut solution. You might consider plotting critical points or in the worst-case scenario, you can just plot some arbitrary points.
A: For the first question, you have to remember the Fundamental Theorem of Calculus, which states:
$$F(x) = \int_a^xf(t)\text{ d}t$$
$$F'(x)=f(x)$$
Obviously the first integral fits this description, so we can say that $g'(x)=f(x)$. Now, to find the extrema of $g(x)$, we have to find a critical point. A critical point is a point where $g'(x)=0$, and we know that $f(x)=g'(x)$, so we can look at the graph of $f(x)$. The function $f(x)=0$ at $x=1$, and by the first derivative test (if a derivative goes from negative to positive at a critical point, that critical point it a minimum), we can tell that $x=1$ is the minimum of $g(x)$. Now, as was stated above, we have to evaluate the function at the endpoints of $f(x)$ to see if $x=1$ is a global minimum. Once you do this, you'll find that $x=1$ is indeed the global minimum of $g(x)$.
For the second question, I would suggest that you remember the rules of curve sketching:


*

*When the first derivative is negative, the function is decreasing.

*When the first derivative is positive, the function is increasing

*Find the maxima of the function (we just did that)

*Plot some points as a guide if necessary

