Question 39

Consider the function: $$f(x) = \mid {2 - \mid x - 1\mid}\mid$$ for all $$x \in R$$. Then the value of $$f'(-2) + f'(0) + f'(2) + f'(4)$$ is

Solution

$$\mid x - 1\mid$$

=$$\begin{cases}x-1 & x \geq 1\\1-x & x < 1 \end{cases}$$

x < 1 f(x)= $$\mid x + 1\mid$$

x $$\geq$$ 1 f(x) = $$\mid 3-x\mid$$

x < -1 f(x) = -x-1

x $$\geq$$ -1 f(x) = x+1

x < 3 f(x) = x-3

x $$\geq$$ 3 f(x) =3-x

Calculate the values of

f'(-2) = -1, f'(0) = 1,f'(2) = 1, f'(4) = -1

$$f'(-2) + f'(0) + f'(2) + f'(4)$$ = 0

Hence B is the correct answer.

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