How do you find the critical points of f when $f (x) = \frac{x^{4}}{5 {(x - 5)}^{2}}$ $f(x)= x^4/(5(x-5)^2)$ ?

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Answer 1 · 2015-05-25T17:22:53

Critical points are x=0, x= 10

First get f'(x)= $\frac{1}{5} [\frac{4 x^{3} {(x - 5)}^{2} - x^{4} 2 (x - 5)}{{(x - 5)}^{4}}]$ $1/5[(4x^3 (x-5)^2 - x^4 2(x-5))/(x-5)^4]$

$\frac{2 x^{3} (x - 5)}{5} [\frac{2 (x - 5) - x}{{(x - 5)}^{4}}]$ $(2x^3 (x-5))/5[(2(x-5)-x)/(x-5)^4]$

$\frac{2 x^{3}}{5} \frac{x - 10}{{(x - 5)}^{3}}$ $(2x^3) /5(x-10)/(x-5)^3$

Solving for f(x)=0, it is x=0, x=10

At x=5, f'(x) does not exist, hence x=5 is also a critical point.

How do you find the critical points of f when f(x)= x^4/(5(x-5)^2)f(x)=x45(x−5)2f(x)= x^4/(5(x-5)^2)?