How do you find the intervals of increasing and decreasing using the first derivative given $y = {(x + 2)}^{2} (x - 1)$ $y=(x+2)^2(x-1)$ ?

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How do you find the intervals of increasing and decreasing using the first derivative given $y = {(x + 2)}^{2} (x - 1)$ $y=(x+2)^2(x-1)$ ?

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Answer 1 · 2018-06-05T13:50:05

Using the product rule:

$\frac{d y}{d x} = 2 (x + 2) (x - 1) + {(x + 2)}^{2} = (x + 2) (2 x - 2 + x + 2) = 3 x (x + 2)$ $dy/dx = 2(x+2)(x-1) +(x+2)^2 = (x+2)(2x-2+x+2) = 3x(x+2)$

So:

$\frac{d y}{d x} < 0$ $dy/dx < 0$ for $x \in (- 2, 0)$ $x in (-2,0)$

$\frac{d y}{d x} > 0$ $dy/dx > 0$ for $\in \in (- \infty, - 2) \cup (0, + \infty)$ $in in (-oo,-2) uu (0,+oo)$

which means that $y (x)$ $y(x)$ is increasing from $- \infty$ $-oo$ to $x = - 2$ $x=-2$ where it has a local maximum, decreasing from $x = - 2$ $x=-2$ to $x = 0$ $x=0$ where it has a local minimum, and again increasing from $x = 0$ $x=0$ to $+ \infty$ $+oo$ .

graph{(x+2)^2(x-1) [-4, 3, -10, 10]}

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How do you find the intervals of increasing and decreasing using the first derivative given $y = {(x + 2)}^{2} (x - 1)$ $y=(x+2)^2(x-1)$ ?

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How do you find the intervals of increasing and decreasing using the first derivative given y=(x+2)^2(x-1)y=(x+2)2(x−1)y=(x+2)^2(x-1)?

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How do you find the intervals of increasing and decreasing using the first derivative given $y = {(x + 2)}^{2} (x - 1)$ $y=(x+2)^2(x-1)$ ?