Question

Find the absolute max and min values of f(x) = x/x^2-x+1 on the interval [0,3]

Answer 1

Have you found f'?

Answer 2

I used quotient rule and got f' (x) = -x^2 + 1/ (x^2+1)^2

Answer 3

\[f(x)=\frac{x}{x^2-x+1} \]
\[f'(x)=\frac{(x)'(x^2-x+1)-x(x^2-x+1)'}{(x^2-x+1)^2}\]
Is this what you did to find f'?

Answer 4

(x)'=1
(x^2-x+1)'=2x-1
Plug those in and simplify the top.

Answer 5

Then find when f' dne and find when f'=0 (think about the domain of f). If the numbers you found from f' dne and f'=0 are numbers that exist in your domain for f, then these are critical numbers for f.
Plug in your critical numbers into f along with your endpoints to find the absolute highest point and absolute lowest point on your graph.

Answer 6

okay I wrote the question wrong the function is f(x) = x/x^2 + 1

Answer 7

so doing quotient rule
f' = 1(x^2+1) - x(2x) / (x^2+1) ^2

Answer 8

so simplified its -x^2 +1 / (x^2+1)^2

Answer 9

now im stuck, how do i find the critical numbers? do I just look at the denominator

Answer 10

when is -x^2+1 equal to 0?
when is (-x^2+1)/(x^2+1)^2 not exist? (it exist everywhere. the bottom is never 0)
So just answer the first question.

Answer 11

when x = 1, or -1

Answer 12

Ok great now your domain for your function was [0,3] and -1 is not included in that so we can throw -1 in the trash but we are keeping 1 since 1 is in [0,3]

Answer 13

Now plug 0,1,3 into your original function

Answer 14

okay

Answer 15

find these
\[f(0), f(1),f(3)\]

Answer 16

f(1) is my abs max since it has the highest value

Answer 17

ok great.

Answer 18

now what is abs min?

Answer 19

f(0)

Answer 20

so quick question, for the denominator part you just look at values to see when it does not exist ? so if it were to have a restriction, then that value would also be a critical number?

Answer 21

yep if that number is in the domain of f.

Answer 22

ok thanks