What is the extreme f(x)=((3x))/((9-x^2)),-3<= x<= 2 ?

The extreme f(x)=((3x))/((9-x^2)),-3<= x<= 2 is Maximum(2, 6/5)

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extreme f(x)=((3x))/((9-x^2)),-3<= x<= 2

extreme points

f (x) = \frac{( 3 x )}{( 9 - x ^{2} )}, - 3 \leq x \leq 2

Maximum (2, \frac{6}{5})

Solution steps

Find the critical points:

x = 2

Domain of

\frac{3 x}{9 - x ^{2}}, - 3 \leq x \leq 2 : - 3 < x \leq 2

Find intervals:

Increasing

: - 3 < x < 2

Plug the extreme point

x = 2

into

\frac{3 x}{9 - x ^{2}} \Rightarrow y = \frac{6}{5}

Maximum (2, \frac{6}{5})

f (x) = \frac{1}{4} x + \frac{1}{x}, x > 0

f (0) = x^{2} - 4 x - 4

\frac{7 e ^{x} + 7 e ^{- x}}{2}

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

f (x) = x^{4} - 72 x^{2} + 1296, - 7 \leq x \leq 7

What is the extreme f(x)=((3x))/((9-x^2)),-3<= x<= 2 ?
The extreme f(x)=((3x))/((9-x^2)),-3<= x<= 2 is Maximum(2, 6/5)