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

The extreme f(x)=((t+3)^3)/((t-1)^2) is Saddle(-3,0),Minimum(9,27)

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

extreme points

f (x) = \frac{( t + 3 ) ^{3}}{( t - 1 ) ^{2}}

Saddle (- 3, 0), Minimum (9, 27)

Solution steps

f^{'} (t) = \frac{( t + 3 ) ^{2} ( t - 9 )}{( t - 1 ) ^{3}}

Find intervals:

Increasing

: - \infty < t < - 3,

Increasing

: - 3 < t < 1,

Decreasing

: 1 < t < 9,

Increasing

: 9 < t < \infty

Plug

t = - 3

into

\frac{( t + 3 ) ^{3}}{( t - 1 ) ^{2}} : 0

Saddle (- 3, 0)

Plug

t = 9

into

\frac{( t + 3 ) ^{3}}{( t - 1 ) ^{2}} : 27

Minimum (9, 27)

Saddle (- 3, 0), Minimum (9, 27)

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

f (x) = 25 x + \frac{16}{x}

f (x) = x^{2} - 2 x - 3, 0 \leq x \leq 1

f (x) = ((x^{2} + y^{2}) e^{\frac{- x}{3}})

f (x) = 7 x + ln (x)

What is the extreme f(x)=((t+3)^3)/((t-1)^2) ?
The extreme f(x)=((t+3)^3)/((t-1)^2) is Saddle(-3,0),Minimum(9,27)