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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!

Evoked
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  • Why have you written that $f'(x)$ is constant? – Sachchidanand Prasad Nov 04 '17 at 12:11
  • Because the exercise said that in the intervals mentioned above w.r.t. the derivative, as you see, they are constants (real numbers), so I think it is just confirming what you can see below (but I just wrote the exercise, so it is what was shown in the exam) @SachchidanandPrasad – Evoked Nov 04 '17 at 12:15

1 Answers1

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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.

Maadhav
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