Is $f (x) = {(- x - 4)}^{2} + 3 x^{2} - 3 x$ $f(x)=(-x-4)^2+3x^2-3x$ increasing or decreasing at $x = - 1$ $x=-1$ ?

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Is $f (x) = {(- x - 4)}^{2} + 3 x^{2} - 3 x$ $f(x)=(-x-4)^2+3x^2-3x$ increasing or decreasing at $x = - 1$ $x=-1$ ?

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Answer 1 · 2016-05-06T14:29:49

$f (x) = {(- x - 4)}^{2} + 3 x^{2} - 3 x$ $f(x)=(-x-4)^2+3x^2-3x$ is decreasing at $x = - 1$ $x=-1$

Explanation:

From the given equation, obtain the first derivative and then substitute the value of $x = - 1$ $x=-1$
$f (x) = {(- x - 4)}^{2} + 3 x^{2} - 3 x$ $f(x)=(-x-4)^2+3x^2-3x$

$f (x) = x^{2} + 8 x + 16 + 3 x^{2} - 3 x$ $f(x)=x^2+8x+16+3x^2-3x$

$f (x) = 4 x^{2} + 5 x + 16$ $f(x)=4x^2+5x+16$

$f'(x)=8x+5$

$f'(-1)=8(-1)+5$

$f'(-1)=-3" "$ this means decreasing

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Is $f (x) = {(- x - 4)}^{2} + 3 x^{2} - 3 x$ $f(x)=(-x-4)^2+3x^2-3x$ increasing or decreasing at $x = - 1$ $x=-1$ ?

1 Answer

Explanation:

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Is f(x)=(-x-4)^2+3x^2-3x f(x)=(−x−4)2+3x2−3xf(x)=(-x-4)^2+3x^2-3x increasing or decreasing at x=-1 x=−1x=-1 ?

1 Answer

Explanation:

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Is $f (x) = {(- x - 4)}^{2} + 3 x^{2} - 3 x$ $f(x)=(-x-4)^2+3x^2-3x$ increasing or decreasing at $x = - 1$ $x=-1$ ?