Prove there is no such c value (mean value theorem)

[olive quest]

did I solve this right? Dont tell me the answer.I already know it.Give me a feedback about my solution.

[dreamy panther]

no. (almost) same Answer as in your other post. The absolute Value function is not differentiable at x=0. Plot the function (e.g. in gnuplot). It has a "kink" at x=0, thus you cannot define a tangent at this point.

the negative x-values don't matter because of the absolute value inside the root

[olive quest]

how can we do tihs witohut graph?

@dreamy panther

@dreamy panther

can you show me this without graphing method?

[warm niche]

it s sqrt(-x)-1\

[thorny jungle]

find the singular point of the given function, which is a composite function of two elementary functions.

[dreamy panther]

The derivative of the absolute value function is the signum function, which is not continuous at x=0

@olive quest

Therefore, abs(x) is not differentiable at x=0

https://study.com/learn/lesson/derivative-absolute-value-function.html

On the other hand: you can write
$f(1) - f(-1) = 0$ \
$f'(c)*2 = 0$ can only be true, if $f'(c)=0$ for any $c \in [-1,1]$. But there is no x with
$f'(x)=0$