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 $f^{\prime}(c)=0$
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