Consider the function f(x)=|x-2|+|x-5|, xR.

Question

Consider the function $f(x)=|x-2|+|x-5|$, $x\in\mathbb{R}$.

Statement 1: $f'(4)=0$

Statement 2: $f$ is continuous in $[2,5]$, differentiable in $(2,5)$ and $f(2)=f(5)$.

Accepted Answer

$$f\left( x \right) = \left| {x - 2} \right| = \left\{ {\begin{pmatrix} {x - 2\,\,\,,} & {x - 2 \ge 0}  \
   {2 - x\,\,\,,} & {x - 2 \le 0}  \

\end{pmatrix} } \right.$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\{ {\begin{pmatrix} {x - 2\,\,\,,} & {x \ge 2}  \
   {2 - x\,\,\,,} & {x \le 2}  \

\end{pmatrix} } \right.$$

Similarly, $$f\left( x \right) = \left| {x - 5} \right| = \left\{ {\begin{pmatrix} {x - 5\,\,\,,} & {x \ge 5}  \
   {5 - x\,\,\,,} & {x \le 5}  \

\end{pmatrix} } \right.$$

$$\therefore$$ $$f\left( x \right) = \left| {x - 2} \right| + \left| {x - 5} \right|$$

$$ = \left\{ {x - 2 + 5 - x = 3,2 \le x \le 5} \right\}$$

Thus $$\,\,\,\,\,\,\,\,\,f\left( x \right) = 3,2 \le x \le 5$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( x \right) = 0,2 < x < 5$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( 4 \right) = 0$$

$$\therefore$$ Statement $1$ is true.

As $$f\left( 2 \right) = 0 + \left| {2 - 5} \right| = 3$$

and $$f\left( 5 \right) = \left| {5 - 2} \right| + 0 = 3$$

$$\therefore$$ Statement - $2$ is also true but not a correct explanation for statement $1.$

Solutions · 1