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

The extreme f(x)=x^4-4x^3+7 is Saddle(0,7),Minimum(3,-20)

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

extreme points

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

Saddle (0, 7), Minimum (3, - 20)

Solution steps

f^{'} (x) = 4 x^{3} - 12 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

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

Saddle (0, 7)

Plug

x = 3

into

x^{4} - 4 x^{3} + 7 : - 20

Minimum (3, - 20)

Saddle (0, 7), Minimum (3, - 20)

(6.7) (- 2.3)

f (x) = \frac{x ^{2}}{x ^{2} + 7}

y + 2 = \frac{5}{2} (x + 1)

(- 4, 1), (2, 4)

f (x) = 10 x + 1

What is the extreme f(x)=x^4-4x^3+7 ?
The extreme f(x)=x^4-4x^3+7 is Saddle(0,7),Minimum(3,-20)