What is the extreme y=(x-1)^3(x-5) ?

The extreme y=(x-1)^3(x-5) is Saddle(1,0),Minimum(4,-27)

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extreme y=(x-1)^3(x-5)

extreme points

y = (x - 1)^{3} (x - 5)

Saddle (1, 0), Minimum (4, - 27)

Solution steps

f^{'} (x) = 4 (x - 1)^{2} (x - 4)

Find intervals:

Decreasing

: - \infty < x < 1,

Decreasing

: 1 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 1

into

(x - 1)^{3} (x - 5) : 0

Saddle (1, 0)

Plug

x = 4

into

(x - 1)^{3} (x - 5) : - 27

Minimum (4, - 27)

Saddle (1, 0), Minimum (4, - 27)

f (x) = 4 x - 6

f (x) = - \frac{2}{x ^{2}}, 0.5 \leq x \leq 2

f (x) = - \frac{2}{x ^{2}}, 0.5 \leq x \leq 5

f (x) = 2 - 3 x

x^{4} - 2 x^{2} - y^{2} + 3

What is the extreme y=(x-1)^3(x-5) ?
The extreme y=(x-1)^3(x-5) is Saddle(1,0),Minimum(4,-27)