Question

How do you find the local max and min for `f (x) = x^(3) - 6x^(2) + 5`?

Answer 1

#

Explanation:

Find the critical values (when `f'(x)=0` or is undefined).

`f'(x)=3x^2-12x`

`f'(x)=3x(x-4)`

`f'(x)=0` when `x=0,4`.

Make a sign chart for `f'(x)`.

`color(white)(xxxxxxxxxxxx)0color(white)(xxxxxxxxxxxx)4`
`larr----------------rarr`
`color(white)(xxx)"POSITIVE"color(white)(xxxxx)"NEGATIVE"color(white)(xxxxx)"POSITIVE"`

There is a relative maximum at `x=0` because the derivative goes from positive to negative.

There is a relative minimum at `x=4` because the derivative goes from negative to positive.

Look at a graph:

graph{x^3-6x^2+5 [-36.65, 55.83, -32.08, 14.18]}

Answer 2

It has a local maxima at `(0, 5)`
It has a local minima at `(4, -27)`

Explanation:

Given -

`y=x^3-6x^2+5`
`dy/dx=3x^2-12x`
`(d^2y)/dx^2=6x-12`

`dy/dx=0 =>3x^2-12x=0`
`3x(x-4)=0`

`3x=0`
`x=0`

`x-4=0`
`x=4`

AT `x=0; (d^2y)/dx^2=6(0)-12=-12 <0`
At `x=0; dy/dx=0;(d^2)/dx^2<0` Hence the function has a local maximum at `x=0`

AT `x=4;(d^2y)/dx^2=6(4)-12=24-12=12>0`

Local Maximum is -
At `x=0; y=(0)^3-6(0)^2+5=5`

`(0, 5)`

AT `x=4; dy/dx=0;(d^2y)/dx^2>0`. Hence the function has a local minimum.

Local Minimum -
At `x=4;y=(4)^3-6(4)^2+5=64-96+5=-27`
`(4, -27)`

graph{x^3-6x^2+5 [-58.5, 58.55, -29.24, 29.3]}

Maxima Minima