What are the local extrema of $f (x) = x^{2} (x + 2)$ $f(x)= x^2(x+2)$ ?

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What are the local extrema of $f (x) = x^{2} (x + 2)$ $f(x)= x^2(x+2)$ ?

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Answer 1 · 2015-12-26T05:59:43

Minima $(0, 0)$ $(0, 0)$
Maxima $(- \frac{4}{3}, 1 \frac{5}{27})$ $(-4/3, 1 5/27)$

Explanation:

Given-

$y = x^{2} (x + 2)$ $y=x^2(x+2)$
$y = x^{3} + 2 x^{2}$ $y=x^3+2x^2$
$\frac{d y}{d x} = 3 x^{2} + 4 x$ $dy/dx=3x^2+4x$
$\frac{d^{2} y}{{d x}^{2}} = 6 x + 4$ $(d^2y)/(dx^2)=6x+4$
$\frac{d y}{d x} = 0 \Rightarrow 3 x^{2} + 4 x = 0$ $dy/dx=0=>3x^2+4x=0$
$x (3 x + 4) = 0$ $x(3x+4)=0$
$x = 0$ $x=0$
$3 x + 4 = 0$ $3x+4=0$
$x = - \frac{4}{3}$ $x=-4/3$
At $x = 0; \frac{d^{2} y}{{d x}^{2}} = 6 (0) + 4 = 4 > 0$ $x=0; (d^2y)/(dx^2)=6(0)+4=4>0$

At $x = 0; \frac{d y}{d x} = 0; \frac{d^{2} y}{{d x}^{2}} > 0$ $x=0; dy/dx=0;(d^2y)/(dx^2)>0$
Hence the function has a minima at $x = 0$ $x=0$

At $x = 0; y = {(0)}^{2} (0 + 2) = 0$ $x=0;y=(0)^2(0+2)=0$
Minima $(0, 0)$ $(0, 0)$

At $x = - \frac{4}{3}; \frac{d^{2} y}{{d x}^{2}} = 6 (- \frac{4}{3}) + 4 = - 4 < 0$ $x=-4/3; (d^2y)/(dx^2)=6(-4/3)+4=-4<0$
At $x = - 4; \frac{d y}{d x} = 0; \frac{d^{2} y}{{d x}^{2}} < 0$ $x=-4; dy/dx=0;(d^2y)/(dx^2)<0$

Hence the function has a maxima at $x = - \frac{4}{3}$ $x=-4/3$

At $x = - \frac{4}{3}; y = {(- \frac{4}{3})}^{2} (- \frac{4}{3} + 2) = 1 \frac{5}{27}$ $x=-4/3;y=(-4/3)^2(-4/3+2)=1 5/27$

Maxima $(- \frac{4}{3}, 1 \frac{5}{27})$ $(-4/3, 1 5/27)$

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What are the local extrema of $f (x) = x^{2} (x + 2)$ $f(x)= x^2(x+2)$ ?

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