What is the extreme f(x)=7x^4-28x^3+5 ?

The extreme f(x)=7x^4-28x^3+5 is Saddle(0,5),Minimum(3,-184)

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extreme f(x)=7x^4-28x^3+5

extreme points

f (x) = 7 x^{4} - 28 x^{3} + 5

Saddle (0, 5), Minimum (3, - 184)

Solution steps

f^{'} (x) = 28 x^{3} - 84 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

7 x^{4} - 28 x^{3} + 5 : 5

Saddle (0, 5)

Plug

x = 3

into

7 x^{4} - 28 x^{3} + 5 : - 184

Minimum (3, - 184)

Saddle (0, 5), Minimum (3, - 184)

f (x) = x^{4} - 2 4^{2} + 180

f (x) = 2 x^{4} - 16 x^{3}

f (x) = x^{2} - 8 x, 0 \leq x \leq 9

f (x) = 2 - 3 x, 0 \leq x \leq 1

f (x) = (\frac{1}{12} x^{3} + \frac{- 1}{2} x^{2} + 12)

What is the extreme f(x)=7x^4-28x^3+5 ?
The extreme f(x)=7x^4-28x^3+5 is Saddle(0,5),Minimum(3,-184)