Let f(x) = 2+|x|-|x-1|+|x+1|, xR.

Question

Let $f(x) = 2+|x|-|x-1|+|x+1|$, $x\in\mathbb{R}$. Consider

(S1): $f'\left(-\dfrac32\right)+f'\left(-\dfrac12\right)+f'\left(\dfrac12\right)+f'\left(\dfrac32\right) = 2$

(S2): $\displaystyle\int_{-2}^2 f(x)\,dx = 12$

Then,

Accepted Answer

Piecewise: $f(x) = \begin{cases}-x, & x<-1\ x+2, & -1\le x<0\ 3x+2, & 0\le x<1\ x+4, & x\ge1\end{cases}$

$$f'\left(-\frac32\right)+f'\left(-\frac12\right)+f'\left(\frac12\right)+f'\left(\frac32\right) = -1+1+3+1 = 4 \ne 2$$

so (S1) is false. For (S2):

$$\int_{-2}^2 f(x)\,dx = \int_{-2}^{-1}(-x)dx+\int_{-1}^0(x+2)dx+\int_0^1(3x+2)dx+\int_1^2(x+4)dx = \frac32+\frac32+\frac72+\frac{11}2 = 12$$

So only (S2) is correct.

Solutions · 1