I have said graph:
Firstly I know that $f$ is an anti-derivative of $g$. So implies $f'(x)=g(x)$, but I am confused what this actually means?
Then I was wondering how to go about these:
1) Identify all the intervals where $f$ is constant.
For this I thought for a function to be constant $f'(x)=0$, so where $g(x)=0$. Which led me to the conclusion that this is $[2,3],[7],[9]$.
2) Identify all values of $x$ where $f$ has a local maximum and all values of $x$ where $f$ has a local minimum. Are these correct?
From the last question I have the points $2,3,7,9$ but I am not sure how to determine if they are minimum or maximum points?
3)Identify all $x$ satisfying $0\lt x \lt 10$ at which the second derivative does not exist.
Then finally I thought places where the second derivative wouldn't exist was where $g$ is constant. So $1,2,3,4,5,6,8$ but again I am unsure?
If anyone could help alleviate my confusion and check my work I would be grateful!
You are given the graph of $g(x)=f'(x)$. What this means is that the graph is giving you information about how $f(x)$ changes. In other words, you are looking at the value of the derivative of $f(x)$ at every point.
1) Your reasoning is correct here. Since $g(x)=f'(x)=0$ on $[2,3]\cup\{7\}\cup\{9\}$, $f$ must be constant there.
2)Your critical points are the $x$ values in part 1. These are the points for which the tangent line is a horizontal line. Consider what happens nearby a locally maximal point. This is a region where the graph of $f$ looks like a hill. To the left of the peak (the critical point), the derivative is positive (the function is increasing until it gets to the peak). To the right, the derivative is negative (the function is decreasing). So for a critical point to be a local maximum you must check that the derivative is positive immediately to the left and negative immediately to the right of the point. There is no such point in your example.
Now, a local minimum is exactly the opposite. If you graphed $f$ (rather than $f'$), it would look like a valley. So you want to check your critical points for a point with a negative derivative to the left and a positive derivative to the right. That's exactly what the derivative does near $x=9$.
3)I am going to assume by the second derivative you mean the second derivative of $f$. Do you remember the definition of a derivative? It's a limit and limits only exist if they agree from the left and the right. Take a look at $x=1$. The derivative ($f''(x)$) is zero to the left and nonzero to the right. So the derivative cannot exist at $x=1$. Try using a similar argument on the points $x=2,3,4,6,8$.