Find the exact location of all th relative and absolute maxima of the function?

Question

F(x)= 2/x - 1/x^2

I have come up with the answer that there is no maximum

Answer 1

First note that the function is not defined at x=0

F'(x) = -2/x^2 + 2/x^3

-2/x^2 + 2/x^3 = 0
2/x^3 = 2/x^2

Multiply both sides by (x^3/2)
1 = x

Now determine if x=1 is a relative max or min

Examine the intervals:
(-infinity, 0): F'(x) is negative
(0,1): F'(x) is positive
(1,infinity): F'(x) is negative


So there is a relative max at x=1

Note there is not a relative min at x=0 because the function is undefined there.

There is no absolute max or min because the function goes off to positive and negative infinity at x=0.

Answer 2

F(x)= 2/x - 1/x^2
F'(x) = -2/x^2 +2/x^3
Set derivative =0 getting -2/x^2+2/x^3 =0
-2x +2=0
x = 1
F(1) = 2-1 =1
The function has a max of 1 at x = 1

Answer 3

F(x) = (2x - 1)/x^2 changing the eqn a little
F'(x) = (1/x^2)(2) + (2x-1)(-2)x^(-3)
= 2/x^2 - 2(2x-1)/x^3 =set=0
x = 2x-1
x = 1 is an extremum
But as x --> 0 from both + and -, F(x) --> infinity
So, you are correct, no max but an asymptote at x = 0

Answer 4

Maxima (and minima) occur when F' = 0.
F(x) = 2x^-1 - x^-2
F'(x) = -2x^-2 + 2x^-3
0 = -2x^-2 + 2x^-3
2x^-2 = 2x^-3
1/x² = 1/x^3
x^3 = x²
which is true for x = 0, where we obviously have an asymptote, and x = 1. so your max is at x = 1. ti-83 graph agrees.

Answer 5

x cant = 0, then fractions would be undifined, and with the help of my trusty TI, i have foun that the horizontal asymtote is x=1, so the limit of the equation is 1, and therefore that must be the maximum value, ah....i love rational functions....yeah im a nerd

Answer 6

u=1/x
f(u)=2u-u^2
f'(u)=0=>2u-2=0
or
u-1=0
substitute u with 1/x
x-1=0
x=1
let's try some numbers:
f(-1)=-3
f(0)= -infinity
f(0.1)=-80
f(0.5)=0
f(1)=1
f(2)=3/4
f(3)=5/9
.
.
.

Answer 7

kyle is right