What is the extreme f(x)=4x^5-7x^2+8 ?

The extreme f(x)=4x^5-7x^2+8 is Maximum(0,8),Minimum(\sqrt[3]{7/10},4(7/10)^{5/3}-7(7/10)^{2/3}+8)

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extreme f(x)=4x^5-7x^2+8

extreme points

f (x) = 4 x^{5} - 7 x^{2} + 8

Solution steps

f^{'} (x) = 20 x^{4} - 14 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

x = 0

into

4 x^{5} - 7 x^{2} + 8 : 8

Maximum (0, 8)

Plug

into

4 x^{5} - 7 x^{2} + 8 : 4 (\frac{7}{10})^{\frac{5}{3}} - 7 (\frac{7}{10})^{\frac{2}{3}} + 8

f (x) = \frac{x}{x ^{2} + 1}, (- 1, 2)

f (x) = 5 x^{3} + x^{4}

f (x) = 9 x^{2} - 4 x

f (x) = - (x + 1)^{2} \cdot (x + 3)^{3} \cdot (x - 4)^{2}

f (x) = 5 e^{x} - e^{2 x}

What is the extreme f(x)=4x^5-7x^2+8 ?
The extreme f(x)=4x^5-7x^2+8 is Maximum(0,8),Minimum(\sqrt[3]{7/10},4(7/10)^{5/3}-7(7/10)^{2/3}+8)