Question

f(x)=3-absolutevalue(x-3) find the derivative.

Answer 1

Rewrite it as a piece-wise function\[
f(x) = 2-|x-3| = \begin{cases}
2-(x-3) &=&-x-1 &x>0\\
2- (-(x-3))& =&x-5 &x<0
\end{cases}\]
Now we can differentiate each piece.
\[
f'(x) = \begin{cases}
(-x-1)' &=&-1 &x>0\\
(x-5)'&=&1 &x<0
\end{cases}
\]

Answer 2

Whoops it should say: \[
f'(x) = \begin{cases}
(-x-1)' &=&-1 &x>3\\
(x-5)'&=&1 &x<3
\end{cases}
\]

Answer 3

\[
f(x) = 2-|x-3| = \begin{cases}
2-(x-3) &=&-x-1 &x-3>0 \implies x>3\\
2- (-(x-3))& =&x-5 &x-3<0 \implies x<3
\end{cases}
\]

Answer 4

I'm trying to determine if I can use Rolle's Theorem for this function... I'm still unsure. Oh and this is on the interval [0,6]

Answer 5

What are you trying to show?

Answer 6

My instructions are: Determine whether Rolle's Theorem can be applied to f on the closed interval [a,b]. If Rolle's Theorem can be applied, find all values of c in the open interval (a,b) such that f'(c)=0.

Answer 7

The problem with Rolle's theorem is that it requires the function be differentiable on the interval, which in this case isn't true. \(f(x)\) is not differentiable at 3.

Answer 8

Basically your interval can't contain \(3\).

Answer 9

Okay, I sort of get it...

Answer 10

Anyway it's pretty obvious that \(f'(x)\neq 0\) for all \(x\) since we know it is either \(-1\) or \(1\).