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Popular >

extreme f(x)=(x^2(x-3))/(x+6)

Solution

extreme points

f (x) = \frac{x ^{2} ( x - 3 )}{x + 6}

Solution

Minimum (\frac{- 15 - 3 57}{4}, \frac{9 ( 19 57 + 143 )}{8}), Maximum (0, 0), Minimum (\frac{- 15 + 3 57}{4}, - \frac{9 ( 19 57 - 143 )}{8})

Solution steps

f^{'} (x) = \frac{2 x ^{3} + 15 x ^{2} - 36 x}{( x + 6 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < \frac{- 15 - 3 57}{4},

Increasing

: \frac{- 15 - 3 57}{4} < x < - 6,

Increasing

: - 6 < x < 0,

Decreasing

: 0 < x < \frac{3 57 - 15}{4},

Increasing

: \frac{3 57 - 15}{4} < x < \infty

Plug

x = \frac{- 15 - 3 57}{4}

into

\frac{x ^{2} ( x - 3 )}{x + 6} : \frac{9 ( 19 57 + 143 )}{8}

Minimum (\frac{- 15 - 3 57}{4}, \frac{9 ( 19 57 + 143 )}{8})

Plug

x = 0

into

\frac{x ^{2} ( x - 3 )}{x + 6} : 0

Maximum (0, 0)

Plug

x = \frac{- 15 + 3 57}{4}

into

\frac{x ^{2} ( x - 3 )}{x + 6} : - \frac{9 ( 19 57 - 143 )}{8}

Minimum (\frac{- 15 + 3 57}{4}, - \frac{9 ( 19 57 - 143 )}{8})

Minimum (\frac{- 15 - 3 57}{4}, \frac{9 ( 19 57 + 143 )}{8}), Maximum (0, 0), Minimum (\frac{- 15 + 3 57}{4}, - \frac{9 ( 19 57 - 143 )}{8})

Popular Examples

extreme f(x)=f(x)=xsqrt(x^2+4)extreme points

f (x) = f (x) = x x^{2} + 4

extreme f(x)=2x^3+3x^2-120x+9 extreme points

f (x) = 2 x^{3} + 3 x^{2} - 120 x + 9

extreme f(x)=x^3+12x^2-27x+6,-10<= x<= 0 extreme points

f (x) = x^{3} + 12 x^{2} - 27 x + 6, - 10 \leq x \leq 0

extreme y=(2x^2-1)^2 extreme points

y = (2 x^{2} - 1)^{2}

extreme f(x)= 6/(-4x+2)extreme points

f (x) = \frac{6}{- 4 x + 2}

Frequently Asked Questions (FAQ)

What is the extreme f(x)=(x^2(x-3))/(x+6) ?
The extreme f(x)=(x^2(x-3))/(x+6) is