Question

How do you find local maximum value of f using the first and second derivative tests: `f(x) = x / (x^2 + 9)`?

Answer 1

The first derivative yields a critical point at `x=3` i.e.
`(df(x))/(dx)=0, " for " x_(1,2)=+-3`
The second derivative `(d^2f(x))/(dx)^2|_(x=3)=-0.01852`

Since `(d^2f(x))/(dx)^2|_(x=3)<0` we have local maximum

Explanation:

Given: `f(x) = x / (x^2 + 9)`

Required: Local Maxima

Solution Strategy:

1) Take the 1st derivative to find the critical point, `c`, by solving
`f'(x)=0`. The solution is the critical point

2) Take the 2nd derivative and evaluate it at `c`:

• If `(d^2f(x))/dx^2|_(x=c)>0; => "local minimum"`
• else `(d^2f(x))/dx^2|_(x=c)<0; => "local maximum"`

1) Use quotient rule to differentiate f(x)

`(df(x))/(dx)=(x^2+9 - 2x^2)/(x^2+9)^2=(9-x^2)/(x^2+9)^2`

Now set `f'(x)=0` and solve

`0=(9-x^2)/(x^2+9)^2` so the critical points are `x_(1,2)=+-3`

2) `(d^2f(x))/dx^2=(-2x(x^2+9)^2- (4x(x^2+9))(9-x^2))/(x^2+9)^4`
`= 2x(x^2-27)/(x^2+9)^3|_(x=3)=-0.01852` Thus we have a maximum