What is the extreme f(x)=t^3-6t^2+9t+1 ?

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

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extreme f(x)=t^3-6t^2+9t+1

extreme points

f (x) = t^{3} - 6 t^{2} + 9 t + 1

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

Solution steps

f^{'} (t) = 3 t^{2} - 12 t + 9

Find intervals:

Increasing

: - \infty < t < 1,

Decreasing

: 1 < t < 3,

Increasing

: 3 < t < \infty

Plug

t = 1

into

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

Maximum (1, 5)

Plug

t = 3

into

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

Minimum (3, 1)

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

\frac{1 - 2 t}{4 + t}

f (x) = - x

f (x) = \frac{10 x ^{2} + 35 x}{49 x ^{2} - 28 x + 4}

f (x) = ln (x^{2} - x - 20)

4 x - 3 y = 12

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