Question

How do you find the local maximum and minimum values of `f(x) = 5 + 9x^2 − 6x^3` using both the First and Second Derivative Tests?

Answer 1

`(0,5) \ \ \ \ \ \ \ =` minimum
`(1,8) \ \ \ \ \ \ \ =` maximum
`(1/2,13/2) =` non-stationary inflection point

Explanation:

We have:

`f(x) = 5 + 9x^2-6x^3`

We can see the critical point via a graph:

graph{5 + 9x^2-6x^3 [-6, 6, -2, 14]}

We can examine the critical points using calculus:

Differentiating wrt `x` we get:

`f'(x) = 18x - 18x^2`

At a critical point we have `f'(x) = 0`

`f'(x) = 0 => 18x - 18x^2 = 0`

`:. 18x(1-x) = 0 => x = 0,1`

And, now we have the `x`-coordinates, we can determine the nature of the turning points (or critical points) by using the second derivative test. Differentiating a second time, we get:

`f''(x) = 18 - 36x`

When:

`x= 0 => f''(0) = 18-0 \ \ gt 0 =>`minimum
`x= 1 => f''(1) = 18-36 lt 0 =>`maximum

Also note we have an inflection point if `f''(x)=0`

`f''(x) = 0 => 18-36x = 0 => x=1/2`

Now we have the `x`-coordinate of the critical points let us find the associated `y`-coordinate:

`x= 0 \ => f(0) \ \ \ \ \ = 5 + 0-0 = 5`
`x= 1 \ => f(1) \ \ \ \ \ = 5 + 9-6 = 8`
`x= 1/2 => f(1/2) = 5 + 9(1/4)-6(1/8) = 13/2`

Hence, in summary

`(0,5) \ \ \ \ \ \ \ =` minimum
`(1,8) \ \ \ \ \ \ \ =` maximum
`(1/2,13/2) =` non-stationary inflection point

Which is consistent with what we see graphically

Answer 2

The local maximum is `=(1,8)` and the local minimum is `=(0,5)`
The point of inflection is `=(1/2,13/2)`

Explanation:

Our function is

`f(x)=5+9x^2-6x^3`

The first derivative is

`f'(x)=18x-18x^2`

The critical points are when

`f'(x)=0`

`18x-18x^2=18x(1-x)`

`18x(1-x)=0`

Therefore, `x=0` and `x=1`

We can build a variation chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `0` `color(white)(aaaaaaa)` `1` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `1-x` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaa)` `↘` `color(white)(aaaa)` `↗` `color(white)(aaaa)` `↘`

Now, we calculate the second derivative

`f''(x)=18-36x`

The point of inflection is when `f''(x)=0`

That is,

`18-36x=0`, `=>`, `x=18/36=1/2`

We build a variation chart with the second derivative

`color(white)(aaaa)` `Interval` `color(white)(aaaa)` `(-oo,1/2)` `color(white)(aaaa)` `(1/2,+oo)`

`color(white)(aaaa)` `sign f''(x)` `color(white)(aaaaaa)` `+` `color(white)(aaaaaaaaaaaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaaaaaa)` `uu` `color(white)(aaaaaaaaaaaa)` `nn`

graph{5+9x^2-6x^3 [-15.35, 16.69, -3.14, 12.88]}