Sketching a continuous function

Question

I had this exercise on an exam and I still don't get it:

Skecth the graph knowing that :

$f(x)$ is continuous at $[-3,3]$

$f'(x)$ is constant

$f(0)=0$

$f'(x)=1$ in $[-3,-1]$

$f'(x)=0$ in $[-1,0]$

$f'(x)=-2$ in $[0,3]$

When I tried the exercise, I started to integrate and substitute the intervals but I got a discountinuous function with the discontinuity points at $f(-1)$ and $f(0)$ because in $[-1,0]$ it is a real number (from my answer).

Where did I go wrong? Any help would be really appreciated!

Answer 1

HINT : Use the constants of integration to remove the discontinuity

• $f(x)=x + C_1$ in $[-3,-1]$

• $f(x)=C_2 $ in $[-1,0]$

• $f(x)=-2x + C_3$ in $[0,3]$

• $C_2$ will be zero and rest can be figured out.