#### Please solve RD Sharma class 12 chapter Mean value theorem exercise 14.1 question 8 sub question (i) maths textbook solution

$f(-2)=f(2),(0,0)$ is the point where tangent will be parallel.

Hint:

$f^{\prime}(x)=2 x$ exist $(-2,2 )$

Given:

$y=x^{2}, x \in[-2,2]$

Explanation:

$y=x^{2}, x \in[-2,2]$

1. Being polynomial$f(x)$ is continuous for all x and hence continuous in$[-2,2]$ .
2. $f^{\prime}(x)=2 x$, which exist $( -2,2)$

$\therefore f(x)$ is derivable in $(-2,2 )$

3.

$\\ f(-2)=(-2)^{2}=4\\\\ f(2)=(2)^{2}=4\\\\ \therefore f(-2)=f(2)$

Therefore, the tangent is parallel to x-axis.