Question

If $f\left(x\right)=2x^2-\left|x\right|+4$ , $x\in [-1,2]$, then for some $c\in (-1,2)$, $f\text{'}\left(c\right)$ is equal to:

• A. $\frac{f\left(2\right)-f\left(0\right)}{2-0}$
• B. $\frac{f\left(2\right)-f\left(-1\right)}{2-\left(-1\right)}$
• C. $\frac{f\left(1\right)-f\left(-1\right)}{1-\left(-1\right)}$
• D. None of these

Answer 1

The correct option is B

$\frac{f\left(2\right)-f\left(-1\right)}{2-\left(-1\right)}$

Explanation for the correct option.

Step 1: Rearrange the given function and find first derivative.

It is given that $f\left(x\right)=2x^2-\left|x\right|+4$ and $x\in [-1,2]$, so

$f\left(x\right)=\left\{\begin{array}{ll}2x^2+x+4,& -1\le x\le 0\\ 2x^2-x+4& 0\le x\le 2\end{array}\right.$

In Case I, where $-1\le x\le 0$,

$$\begin{gathered} f\left(c\right)=2c^2+c+4 \\ \Rightarrow f\text{'}\left(c\right)=4c+1...\left(1\right) \end{gathered}$$

In Case II, where $0\le x\le 2$,

$$\begin{gathered} f\left(c\right)=2c^2-c+4 \\ \Rightarrow f\text{'}\left(c\right)=4c-1...\left(2\right) \end{gathered}$$

Step 2: Apply Lagrange’s Mean Value Theorem.

By Lagrange’s Mean Value Theorem

$f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

From $\left(1\right)$, we get

$\begin{array}{rcl}4c+1& =& \frac{f\left(0\right)-f(-1)}{0+1}\\ & \Rightarrow & 4c+1=\frac{2{\left(0\right)}^2+\left(0\right)+4-\left[2{\left(-1\right)}^2+\left(-1\right)+4\right]}{1}\\ & \Rightarrow & 4c+1=-1\\ \Rightarrow c& =& -\frac{1}{2}\end{array}$

Here, $c\in (-1,0)$.

From $\left(2\right)$, we get

$\begin{array}{rcl}4c-1& =& \frac{f\left(2\right)-f\left(0\right)}{2-0}\\ & \Rightarrow & 4c+1=\frac{2{\left(2\right)}^2+\left(2\right)+4-\left[2{\left(0\right)}^2-\left(0\right)+4\right]}{2}\\ & \Rightarrow & 4c+1=5\\ \Rightarrow c& =& 1\end{array}$

Here, $c\in (0,2)$.

Therefore, for some $c\in (-1,2)$, $f\text{'}\left(c\right)$ is equal to $\frac{f\left(2\right)-f\left(-1\right)}{2-\left(-1\right)}$.

Hence, option B is correct.