Finding Absolute Extrema on a Closed Interval

silverbomb20
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Homework Statement


FIND THE ABSOLUTE MINIMUM AND ABSOLUTE MAXIMUM OF:

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

Homework Equations





The Attempt at a Solution



I don't know how to get started!
 
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hi silverbomb20 - any ideas or thoughts on what defines a maxima or minima?
 
i know i need to use the power rule to get it started, but i don't 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
 
Yes, d/dx(xn) = nxn-1. And 1/x = x-1, so you can use the power rule on that.
 
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...
 
okay so would it be 9 + -1x^-2 ?

im a history major having a lot of trouble with calculus...please help me
 
yes, that looks correct

so when does

9-\frac{1}{x^2} = 0 ?
 
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 doesn't factor
 
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
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
okay so that's going to give me 9x^2=x^2?

or does that -1x^-2 not cancel?
 
  • #12
silverbomb20 said:
okay so that's going to give me 9x^2=x^2?
No. Try again.
silverbomb20 said:
or does that -1x^-2 not cancel?
 
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