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

The extreme f(x)=(x^5)/5-(4x^4)/4+2 is Maximum(0,2),Minimum(4,-246/5)

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

extreme

Maximum (0, 2), Minimum (4, - \frac{246}{5})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

\frac{x ^{5}}{5} - \frac{4 x ^{4}}{4} + 2 : 2

Maximum (0, 2)

Plug

x = 4

into

\frac{x ^{5}}{5} - \frac{4 x ^{4}}{4} + 2 : - \frac{246}{5}

Minimum (4, - \frac{246}{5})

Maximum (0, 2), Minimum (4, - \frac{246}{5})

What is the extreme f(x)=(x^5)/5-(4x^4)/4+2 ?
The extreme f(x)=(x^5)/5-(4x^4)/4+2 is Maximum(0,2),Minimum(4,-246/5)