What is the extreme f(x)=x^4-8x^3+18x^2-16x+40 ?

The extreme f(x)=x^4-8x^3+18x^2-16x+40 is Saddle(1,35),Minimum(4,8)

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extreme f(x)=x^4-8x^3+18x^2-16x+40

extreme points

f (x) = x^{4} - 8 x^{3} + 18 x^{2} - 16 x + 40

Saddle (1, 35), Minimum (4, 8)

Solution steps

f^{'} (x) = 4 x^{3} - 24 x^{2} + 36 x - 16

Find intervals:

Decreasing

: - \infty < x < 1,

Decreasing

: 1 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 1

into

x^{4} - 8 x^{3} + 18 x^{2} - 16 x + 40 : 35

Saddle (1, 35)

Plug

x = 4

into

x^{4} - 8 x^{3} + 18 x^{2} - 16 x + 40 : 8

Minimum (4, 8)

Saddle (1, 35), Minimum (4, 8)

3 x^{2} - x y + 5 y^{2}

f (x) = - x^{3} - 9 x^{2} + 4

x^{2} + 1 + e^{x y}

f (x, y) = y + 81 - x^{2} - y^{2}

f (x) = \frac{2 x ^{3}}{3} + \frac{7 x ^{2}}{2} - 4 x

What is the extreme f(x)=x^4-8x^3+18x^2-16x+40 ?
The extreme f(x)=x^4-8x^3+18x^2-16x+40 is Saddle(1,35),Minimum(4,8)