Absolute minimum and maximum

1. Mar 18, 2009

silverbomb20

1. The problem statement, all variables and given/known data
FIND THE ABSOLUTE MINIMUM AND ABSOLUTE MAXIMUM OF:

f(x) = 9x + 1/x
on the interval [1,3]

2. Relevant equations

3. The attempt at a solution

I dont know how to get started!!!

2. Mar 18, 2009

lanedance

hi silverbomb20 - any ideas or thoughts on what defines a maxima or minima?

3. Mar 18, 2009

silverbomb20

i know i need to use the power rule to get it started, but i dont know how to apply it to the fraction..

derivative of x ^n is n * x^(n-1) right?

i know the 9x would go to just 9
but i am unsure about the fraction

4. Mar 18, 2009

Staff: Mentor

Yes, d/dx(xn) = nxn-1. And 1/x = x-1, so you can use the power rule on that.

5. Mar 18, 2009

lanedance

first part sounds good

so we're stuck on
$$\frac{d}{dx} (\frac{1}{x})$$

you could write it like below and use the power rule
$$\frac{d}{dx} (x^{-1})$$

or you could try and do it from first principles...

6. Mar 18, 2009

silverbomb20

okay so would it be 9 + -1x^-2 ?

7. Mar 18, 2009

lanedance

yes, that looks correct

so when does

$$9-\frac{1}{x^2} = 0$$ ?

8. Mar 19, 2009

silverbomb20

in my mind...never.

f1(x) = 9 + -1x^-2

so i have to set it equal to zero right?

so 9 + -1x^-2=0

but that doesnt factor

9. Mar 19, 2009

Staff: Mentor

Multiply both sides by x^2.

Before we get (more) bogged down with minutiae, let's look at the strategy. Extreme values of a function f will happen at values of x for which f'(x) is zero, or at endpoints of the interval of definition.

10. Mar 19, 2009

lanedance

f'(x) represents a critical point which could be a maxima, minima or point of inflection, but you nee dto check to find which

as we are talking about absolute max & min we also need to check the boundaries of [1,3] ie the points x=1 and x=3

as for the local minima you're on your way
9 + -1x^-2=0

try first multiplying both sides of the equation by x^2

11. Mar 19, 2009

silverbomb20

okay so thats gonna give me 9x^2=x^2?

or does that -1x^-2 not cancel?

12. Mar 19, 2009

Staff: Mentor

No. Try again.