Consider the function f ( x ) = 3 x 2 x 2 − 9 f\left(x\right)=\frac{3x^2}{x^2-9} on the interval [ − 5 , 5 ] \left\lbrack-5,5\right\rbrack . Find the critical points of f f and use the First Derivative Test to classify these points. Then, determine the absolute maximum and minimum values of f f on the specified interval (if there are any).

Critical point: x = 0 x=0 x = 0 Local maximum at x = 0 x=0 x = 0 No absolute extrema

Critical point: x = 0 x=0 x = 0 Local minimum at x = 0 x=0 x = 0 No absolute extrema

Critical points: x = − 3 x=-3 , x = 0 x=0 x = 0 , and x = 3 x=3 Local minimum at x = − 3 x=-3 and local maximum at x = 3 x=3 Absolute minimum: 2 2 at x = 1 x=1 Absolute maximum: 4 4 at x = − 1 x=-1

Critical points: x = − 3 x=-3 x = − 3 , x = 0 x=0 x = 0 , and x = 3 x=3 x = 3 Local minimum at x = − 3 x=-3 x = − 3 and local maximum at x = 3 x=3 x = 3 Absolute minimum: 2 2 2 at x = − 1 x=-1 Absolute maximum: 4 4 4 at x = 1 x=1

Consider the function f(x)=3x2x2−9f\left(x\right)=...

A

Critical point: $x=0$

Local minimum at $x=0$

No absolute extrema

B

Critical point: $x=0$

Local maximum at $x=0$

No absolute extrema

C

Critical points: $x = - 3$ , $x=0$ , and $x equals 3$

Local minimum at $x = - 3$ and local maximum at $x equals 3$

Absolute minimum: $2$ at $x equals 1$

Absolute maximum: $4$ at $x = - 1$

D

Critical points: $x=-3$ , $x=0$ , and $x=3$

Local minimum at $x=-3$ and local maximum at $x=3$

Absolute minimum: $2$ at $x = - 1$

Absolute maximum: $4$ at $x equals 1$

The First Derivative Test: Study with Video Lessons, Practice Problems & Examples