#### Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 7 sub question (i)

Geometrical interpretation of Rolle’s Theorem, (0,16) is the point on curve$y = 16 - x ^2$ , where tangent is parallel to $x-$axis.

Hint:

All conditions of Rolle’s Theorem are satisfied.

Given:

$y=16-x^{2}, x \in[-1,1]$

Explanation:

$y=16-x^{2}, x \in[-1,1]$

1. Being polynomial f(x) is continuous for all x and hence continuous in$[-1,1]$.
2. f(x) is derivable in $(-1,1)$
3. $f(-1)=15=f(1)$

Thus, all conditions of Rolle’s Theorem are satisfied.

Therefore, there exists at least one$c \in(-1,1)$ such that$f'(c) = 0$

Let’s find the derivative of f(x)

$\Rightarrow \quad f^{\prime}(x)=-2 x\) \ \ \ \ \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right] \\\\$

Now,

$$$f^{\prime}(c)=0$$ \\\\ \Rightarrow \quad-2 c=0 \\\\ \Rightarrow \quad c=0$

Also,      $f(0) = 16 -0 = 16$

Hence by geometrical interpretation of Rolle’s Theorem (0,0)is the point on curve$y = 16 - x^2$, where tangent is parallel to x-axis.