Verify Rolle's theorem for the function $y=x^{2}+2, a=-2$ and $b=2$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

The function $y=x^{2}+2$ is continuous in $[-2,2]$ and differentiable in $(-2,2).$

Also $f(-2)=f(2)=6$ and hence the value of $f(x)$ at $-2$ and $2$ coincide. Rolle's theorem states that there is a point $c \in(-2,2),$ where $f^{\prime}(c)=0 .$ Since $f^{\prime}(x)=2 x,$ we get $c=0 .$ Thus at $c=0,$ we have $f^{\prime}(c)=0$ and $c=0 \in(-2,2)$

Similar Questions

Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then

If Rolle's theorem holds for the function $f(x) = 2{x^3} + b{x^2} + cx,\,x\, \in \,\left[ { - 1,1} \right]$ at the point $x = \frac{1}{2}$ , then $(2b+c)$ is equal to 

In which of the following functions is Rolle's theorem applicable ?

Verify Mean Value Theorem, if $f(x)=x^{2}-4 x-3$ in the interval $[a, b],$ where $a=1$ and $b=4$

Functions $f(x)$ and $g(x)$ are such that $f(x) + \int\limits_0^x {g(t)dt = 2\,\sin \,x\, - \,\frac{\pi }{2}} $ and $f'(x).g (x) = cos^2\,x$ , then number of solution $(s)$ of equation $f(x) + g(x) = 0$ in $(0,3 \pi$) is-