Verify Mean Value Theorem, if $f(x)=x^{3}-5 x^{2}-3 x$ in the interval $[a, b],$ where $a=1$ and $b=3 .$ Find all $c \in(1,3)$ for which $f^{\prime}(c)=0$
Verify Mean Value Theorem, if $f(x)=x^{3}-5 x^{2}-3 x$ in the interval $[a, b],$ where $a=1$ and $b=3 .$ Find all $c \in(1,3)$ for which $f^{\prime}(c)=0$
The given function $f$ is $f(x)=x^{2}-5 x^{2}-3 x$
$f,$ being a polynomial function, is continuous in $[1,3],$ and is differentiable in $(1,3)$
Whose derivative is $3 x^{2}-10 x-3$
$f(1)=1^{2}-5 \times 1^{2}-3 \times 1=-7, f(3)=3^{3}-3 \times 3=27$
$\therefore \frac{f(b)-f(a)}{b-a}=\frac{f(3)-f(1)}{3-1}=\frac{-27-(-7)}{3-1}=-10$
Mean Value Theorem states that there exist a point $c \in(1,3)$ such that $f^{\prime}(c)=-10$
$f^{\prime}(c)=-10$
$\Rightarrow 3 c^{2}-10 c-3=10$
$\Rightarrow 3 c^{2}-10 c+7=0$
$\Rightarrow 3 c^{2}-3 c-7 c+7=0$
$\Rightarrow 3 c(c-1)-7(c-1)=0$
$\Rightarrow(c-1)(3 c-7)=0$
$\Rightarrow c=1, \frac{7}{3}$ where $c=\frac{7}{3} \in(1,3)$
Hence, Mean Value Theorem is verified for the given function and $c=\frac{7}{3} \in(1,3)$ is the only point for which $f^{\prime}(c)=0$
Similar Questions
If the function $f(x) = {x^3} - 6a{x^2} + 5x$ satisfies the conditions of Lagrange's mean value theorem for the interval $[1, 2] $ and the tangent to the curve $y = f(x) $ at $x = {7 \over 4}$ is parallel to the chord that joins the points of intersection of the curve with the ordinates $x = 1$ and $x = 2$. Then the value of $a$ is
If the function $f(x) = {x^3} - 6a{x^2} + 5x$ satisfies the conditions of Lagrange's mean value theorem for the interval $[1, 2] $ and the tangent to the curve $y = f(x) $ at $x = {7 \over 4}$ is parallel to the chord that joins the points of intersection of the curve with the ordinates $x = 1$ and $x = 2$. Then the value of $a$ is
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
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
- [JEE MAIN 2014]
Given $f (x) =4\,\, - \,\,{\left( {\frac{1}{2}\, - \,x} \right)^{2/3}}\,$ $g (x) = \left\{ \begin{array}{l}\frac{{\tan \,\,[x]}}{x}\,\,\,\,,\,\,x \ne \,0\\1\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x\, = \,0\end{array} \right.$
$h (x) = \{x\}$ $k (x) = {5^{{{\log }_2}(x\, + \,3)}}$then in $[0, 1]$ Lagranges Mean Value Theorem is $NOT$ applicable to
Given $f (x) =4\,\, - \,\,{\left( {\frac{1}{2}\, - \,x} \right)^{2/3}}\,$ $g (x) = \left\{ \begin{array}{l}\frac{{\tan \,\,[x]}}{x}\,\,\,\,,\,\,x \ne \,0\\1\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x\, = \,0\end{array} \right.$
$h (x) = \{x\}$ $k (x) = {5^{{{\log }_2}(x\, + \,3)}}$then in $[0, 1]$ Lagranges Mean Value Theorem is $NOT$ applicable to
Let $f(x) = (x-4)(x-5)(x-6)(x-7)$ then -
Let $f(x) = (x-4)(x-5)(x-6)(x-7)$ then -
Consider the function $f (x) = 8x^2 - 7x + 5$ on the interval $[-6, 6]$. The value of $c$ that satisfies the conclusion of the mean value theorem, is
Consider the function $f (x) = 8x^2 - 7x + 5$ on the interval $[-6, 6]$. The value of $c$ that satisfies the conclusion of the mean value theorem, is