Question

The value of $c$ in mean value theorem for the function $f\left(x\right)=2x^2+3x+4$ in the interval $\left[1,2\right]$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{3}{2}$
• D. $\frac{2}{3}$

Answer 1

The correct option is C

$\frac{3}{2}$

Explanation for the correct option:

Find the value of $c$.

A function $f\left(x\right)=2x^2+3x+4$ is given.

$f\text{'}\left(x\right)=4x+3$

LaGrange's mean value theorem states that if a function $g\left(x\right)$ is continuous and differentiable in $[a,b]$ then there exists a value $c\in [a,b]$ for which $f\text{'}\left(c\right)=\frac{f\left(a\right)-f\left(b\right)}{a-b}$.

Here $[a,b]$ $=\left[1,2\right]$

So,

$$\begin{gathered} 4c+3=\frac{2+3+4-8-6-4}{1-2} \\ \Rightarrow 4c+3=\frac{-9}{-1} \\ \Rightarrow 4c+3=9 \\ \Rightarrow 4c=6 \\ \Rightarrow c=\frac{3}{2}\in [1,2] \end{gathered}$$

Therefore, the required value of $c$ is $\frac{3}{2}$.

Hence, option $C$ is the correct answer.