# 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?

##### 2 Answers

# (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

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

At a critical point we have

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

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

And, now we have the

# 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 => 18-36x = 0 => x=1/2 #

Now we have the

# 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

The local maximum is

The point of inflection is

#### Explanation:

Our function is

The first derivative is

The critical points are when

Therefore,

We can build a variation chart

Now, we calculate the second derivative

The point of inflection is when

That is,

We build a variation chart with the second derivative

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