What is the extreme f(x)=9x^4-26x^3-84x^2+72 ?

The extreme f(x)=9x^4-26x^3-84x^2+72 is Minimum(-4/3 , 344/27),Maximum(0,72),Minimum(7/2 ,-11539/16)

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extreme f(x)=9x^4-26x^3-84x^2+72

extreme points

f (x) = 9 x^{4} - 26 x^{3} - 84 x^{2} + 72

Minimum (- \frac{4}{3}, \frac{344}{27}), Maximum (0, 72), Minimum (\frac{7}{2}, - \frac{11539}{16})

Solution steps

f^{'} (x) = 36 x^{3} - 78 x^{2} - 168 x

Find intervals:

Decreasing

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

Increasing

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

Decreasing

: 0 < x < \frac{7}{2},

Increasing

: \frac{7}{2} < x < \infty

Plug

x = - \frac{4}{3}

into

9 x^{4} - 26 x^{3} - 84 x^{2} + 72 : \frac{344}{27}

Minimum (- \frac{4}{3}, \frac{344}{27})

Plug

x = 0

into

9 x^{4} - 26 x^{3} - 84 x^{2} + 72 : 72

Maximum (0, 72)

Plug

x = \frac{7}{2}

into

9 x^{4} - 26 x^{3} - 84 x^{2} + 72 : - \frac{11539}{16}

Minimum (\frac{7}{2}, - \frac{11539}{16})

Minimum (- \frac{4}{3}, \frac{344}{27}), Maximum (0, 72), Minimum (\frac{7}{2}, - \frac{11539}{16})

f (x, y) = x^{3} + y^{2} - 3 x - 4 y

\frac{x ^{2} - 11}{x + 6}

f (x, y) = x^{3} - 12 x - 3 y + y^{3}

f (x) = - x^{3} + 3 x + 4

f (x) = \frac{x ^{2} - 8}{x + 3}

What is the extreme f(x)=9x^4-26x^3-84x^2+72 ?
The extreme f(x)=9x^4-26x^3-84x^2+72 is Minimum(-4/3 , 344/27),Maximum(0,72),Minimum(7/2 ,-11539/16)