Answer
$ f'( 1)=f'(2)=0$
$ f'( 4)=\frac{1}{2}$
$ f'( 7) =0$
Work Step by Step
From the given figure, we have:
For $ 0\leq x\leq 3$, the slope of the line tangent to the graph of $f(x)$ is a horizontal line, so $$ f'( 1)=f'(2)=0$$
For $ 3\leq x\leq 5$, the slope of the line tangent to the graph of $f(x)$ is given by
$$\frac{f(5)-f(3)}{5-3}=\frac{1}{2}$$
Then $$ f'( 4)=\frac{1}{2}$$
For $ 5\leq x\leq 9$, the slope of the line tangent to the graph of $f(x)$ is a horizontal line, so $$ f'( 7) =0$$
