Verify Rolle's theorem for the function $y=x^{2}+2, a=-2$ and $b=2$
Verify Rolle's theorem for the function $y=x^{2}+2, a=-2$ and $b=2$
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
Rolle's theorem is true for the function $f(x) = {x^2} - 4 $ in the interval
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Which of the following function can satisfy Rolle's theorem ?
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Let $f(x) = (x-4)(x-5)(x-6)(x-7)$ then -
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If the Rolle's theorem holds for the function $f(x) = 2x^3 + ax^2 + bx$ in the interval $[-1, 1 ]$ for the point $c = \frac{1}{2}$ , then the value of $2a + b$ is
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- [JEE MAIN 2014]
Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
{{x^2}\ln x,\,x > 0} \\
{0,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0}
\end{array}} \right\}$, Rolle’s theorem is applicable to $ f $ for $x \in [0,1]$, if $\alpha = $
Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
{{x^2}\ln x,\,x > 0} \\
{0,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0}
\end{array}} \right\}$, Rolle’s theorem is applicable to $ f $ for $x \in [0,1]$, if $\alpha = $
- [IIT 2004]