What is the extreme f(x)=-1/4 x^4+x^3+5 ?

The extreme f(x)=-1/4 x^4+x^3+5 is Saddle(0,5),Maximum(3, 47/4)

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extreme f(x)=-1/4 x^4+x^3+5

extreme points

f (x) = - \frac{1}{4} x^{4} + x^{3} + 5

Saddle (0, 5), Maximum (3, \frac{47}{4})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Increasing

: 0 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = 0

into

- \frac{1}{4} x^{4} + x^{3} + 5 : 5

Saddle (0, 5)

Plug

x = 3

into

- \frac{1}{4} x^{4} + x^{3} + 5 : \frac{47}{4}

Maximum (3, \frac{47}{4})

Saddle (0, 5), Maximum (3, \frac{47}{4})

f (x) = x^{2} + 8 x + 16

f (x) = e^{2 x}

f (x, y) = x y + \frac{4}{x} + \frac{2}{y}

x^{2} - 7 x + 7.7

f (x) = 2 x + 4

What is the extreme f(x)=-1/4 x^4+x^3+5 ?
The extreme f(x)=-1/4 x^4+x^3+5 is Saddle(0,5),Maximum(3, 47/4)