What is the extreme y=x^3-6x^2+9x+1 ?

The extreme y=x^3-6x^2+9x+1 is Maximum(1,5),Minimum(3,1)

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extreme y=x^3-6x^2+9x+1

extreme points

y = x^{3} - 6 x^{2} + 9 x + 1

Maximum (1, 5), Minimum (3, 1)

Solution steps

f^{'} (x) = 3 x^{2} - 12 x + 9

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 1

into

x^{3} - 6 x^{2} + 9 x + 1 : 5

Maximum (1, 5)

Plug

x = 3

into

x^{3} - 6 x^{2} + 9 x + 1 : 1

Minimum (3, 1)

Maximum (1, 5), Minimum (3, 1)

f (x) = \frac{x ^{2}}{x ^{2} + 48}

f (x, y) = ln (3 - 4 x^{2} - 5 y^{2})

y = x^{2} + 9 x - 2

f (x) = 5 cos^{2} (x), (0, π)

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

What is the extreme y=x^3-6x^2+9x+1 ?
The extreme y=x^3-6x^2+9x+1 is Maximum(1,5),Minimum(3,1)