#### Explain solution for  RD Sharma class 12 chapter Mean value theorem exercise 14.1 question 3 sub question (xvi) math

$c = 2.5 \in (1,4)$

Hint:

$f(1) = f(4)$ , so there exists at$c \in (1,4)$

Given:

$f(x)=x^{2}-5 x+4$ on$[1,4]$

Explanation:

$f(x)=x^{2}-5 x+4$ on$[1,4]$

Since given function f(x) is a polynomial, it is continuous and differentiable everywhere i.e. on R.

Let us find the values at extreme

$f(1)=1^{2}-5(1)+4 \\\\ \therefore \quad f(1)=1-5+4=-4+4 \\\\ \therefore \quad f(1)=0 \\\\ f(4)=4^{2}-5(4)+4 \\\\ f(4)=16-20+4=-4+4 \\\\ \therefore \quad f(4)=0$

We have$f(1)=f(4)$, so there exists at$c \in (1,4)$ , such that$f' (c) = 0$

Let’s find the derivative of$f (x)$

$\begin{array}{l} \Rightarrow \quad f^{\prime}(x)=\frac{d\left(x^{2}-5 x+4\right)}{d x} \\\\ \Rightarrow \quad f^{\prime}(x)=2 x-5 \ \ \ \ \ \ \ {\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]} \end{array}$

We have$f' (c) = 0$

$\\ \Rightarrow 2c- 5 = 0 \\\\ \Rightarrow 2c = 5 \\\\ \Rightarrow c = \frac{5}{2} \\\\ \Rightarrow c = 2.5 \in (1,4)$

Hence Theorem is verified.