What is the extreme f(x)=y=(3x^2)/(x+1),-1<= x<= 5 ?

The extreme f(x)=y=(3x^2)/(x+1),-1<= x<= 5 is Minimum(0,0),Maximum(5, 25/2)

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extreme f(x)=y=(3x^2)/(x+1),-1<= x<= 5

extreme points

f (x) = y = \frac{3 x ^{2}}{x + 1}, - 1 \leq x \leq 5

Minimum (0, 0), Maximum (5, \frac{25}{2})

Solution steps

Find the critical points:

x = 0, x = 5

Domain of

\frac{3 x ^{2}}{x + 1}, - 1 \leq x \leq 5 : - 1 < x \leq 5

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 5

Plug the extreme point

x = 0

into

\frac{3 x ^{2}}{x + 1} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 5

into

\frac{3 x ^{2}}{x + 1} \Rightarrow y = \frac{25}{2}

Maximum (5, \frac{25}{2})

Minimum (0, 0), Maximum (5, \frac{25}{2})

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

f (x) = \frac{x}{ln ( 9 x )}

1 x^{3} - 4 x^{2} - 3 x + 30

What is the extreme f(x)=y=(3x^2)/(x+1),-1<= x<= 5 ?
The extreme f(x)=y=(3x^2)/(x+1),-1<= x<= 5 is Minimum(0,0),Maximum(5, 25/2)