Let $f(x) = (x - 7)^{2}(x - 2)^{7},x \in \lbrack 2,7 brack.\$The value of $ heta \in (2,7)$ such that $f^{'}( heta) = 0$ is equal to

Question

Let $f(x) = (x - 7)^{2}(x - 2)^{7},x \in \lbrack 2,7brack.\$The value of $	heta \in (2,7)$ such that $f^{'}(	heta) = 0$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Given, $f(x) = (x - 7)^{2}(x - 2)^{7}$
$\Rightarrow \ \ \ \ \ \ \ \ f(	heta) = (	heta - 7)^{2}(	heta - 2)^{7}$
On differentiating w.r.t. $	heta,\ we\ get\$
$\Rightarrow \ \ \ \ \ \ \ f^{'}(	heta) = 2(	heta - 7)(	heta - 2)^{7} + 7(	heta - 2)^{6}(	heta - 7)^{2}$
put $f^{'}(	heta) = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ (	heta - 7)(	heta - 2)^{6}\left\lbrack 2(	heta - 2) + 7(	heta - 7) ightbrack = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9	heta$= $53 \Rightarrow 	heta = \frac{53}{9}$

Q. Let $f(x) = (x - 7)^{2}(x - 2)^{7},x \in \lbrack 2,7\rbrack.\$The value of $\theta \in (2,7)$ such that $f^{'}(\theta) = 0$ is equal to

Explanation

Reviewed by Dr. Abubacker

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