Is $f (x) = \frac{- x^{3} - 2 x^{2} - 12 x + 2}{x - 4}$ $f(x)=(-x^3-2x^2-12x+2)/(x-4)$ increasing or decreasing at $x = 3$ $x=3$ ?

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Is $f (x) = \frac{- x^{3} - 2 x^{2} - 12 x + 2}{x - 4}$ $f(x)=(-x^3-2x^2-12x+2)/(x-4)$ increasing or decreasing at $x = 3$ $x=3$ ?

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Answer 1 · 2017-12-07T14:05:35

$f (x)$ $f(x)$ is increasing at $x = 3$ $x=3$

Explanation:

Simplifying $\frac{- x^{3} - 2 x^{2} - 12 x + 2}{x - 4}$ $(-x^3-2x^2-12x+2)/(x-4)$
by synthetic division:
${: (," | ",color(gray)(x^3),color(grey)(x^2),color(gray)(x^1),color(grey)(x^0)), (," | ",-1,-2,-12,+2), (ul(+color(white)("xx"))," | ",ul(color(white)("XX")),ul(-4),ul(-24),ul(-124)), (xx4," | ",-1,-6,-36,-126), (,,color(gray)(x^2),color(gray)(x^1),color(gray)(x^0),color(gray)(x^(-1))) :}$

$f(x)=-x^2-6x-36-126x^(-1)$

$f'(x)=2x-6color(white)("xx")+126x^(-2)$

$f'(3) = -6 -6 +126/(3^2)$
$color(white)("XXX")=12+14$
$color(white)("XXX")=+2$

Since $f'(x) > 0$ at $x=3$
$f(x)$ is increasing at $x=3$

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Is $f (x) = \frac{- x^{3} - 2 x^{2} - 12 x + 2}{x - 4}$ $f(x)=(-x^3-2x^2-12x+2)/(x-4)$ increasing or decreasing at $x = 3$ $x=3$ ?

1 Answer

Explanation:

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Is f(x)=(-x^3-2x^2-12x+2)/(x-4)f(x)=−x3−2x2−12x+2x−4f(x)=(-x^3-2x^2-12x+2)/(x-4) increasing or decreasing at x=3x=3x=3?

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Explanation:

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Is $f (x) = \frac{- x^{3} - 2 x^{2} - 12 x + 2}{x - 4}$ $f(x)=(-x^3-2x^2-12x+2)/(x-4)$ increasing or decreasing at $x = 3$ $x=3$ ?