What are the local extrema of $f (x) = \frac{(x - 2) {(x - 4)}^{3}}{x^{2} - 2}$ $f(x)= ((x-2)(x-4)^3)/(x^2-2)$ ?

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What are the local extrema of $f (x) = \frac{(x - 2) {(x - 4)}^{3}}{x^{2} - 2}$ $f(x)= ((x-2)(x-4)^3)/(x^2-2)$ ?

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Answer 1 · 2017-08-20T13:24:54

$X = - 5$ $X=-5$

Explanation:

$f (x) = \frac{(x - 2) {(x - 4)}^{3}}{x^{2} - 2}$ $f(x)=[(x-2)(x-4)^3]/(x^2-2)$
$x^{2} - 2 = (x + 2) (x - 2)$ $x^2-2=(x+2)(x-2)$
So the function will became:
$f (x) = \frac{{(x - 4)}^{3}}{x + 2}$ $f(x)=[(x-4)^3]/(x+2)$
Now
$f'(x)=d/dx[(x-4)^3]/(x+2)$
$f'(x)=[3(x+2)(x-4)^2-(x-4)^3]/(x+2)^2$
For local extremum point
$f'(x)=0$
So
$[3(x+2)(x-4)^2-(x-4)^3]/(x+2)^2=0$
$[3(x+2)(x-4)^2-(x-4)^3]=0$
$3(x+2)(x-4)^2=(x-4)^3$
$3x+6=x-4$
$2x=-10$
$x=-5$

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What are the local extrema of $f (x) = \frac{(x - 2) {(x - 4)}^{3}}{x^{2} - 2}$ $f(x)= ((x-2)(x-4)^3)/(x^2-2)$ ?

1 Answer

Explanation:

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Explanation:

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What are the local extrema of $f (x) = \frac{(x - 2) {(x - 4)}^{3}}{x^{2} - 2}$ $f(x)= ((x-2)(x-4)^3)/(x^2-2)$ ?