Question

can someone help me find the absolute and local extrema in this equation (3x-4)/(x^2+1)

Answer 1

sure

Answer 2

(x^2+1)(3) - (3x-4)(2x) = 0

Answer 3

on the intervals [-2,2]

Answer 4

3x^2 +3 -6x^2 +8x = 0

-3x^2 +3 +8x = 0

0 = 3x^2 -8x -3

Answer 5

(x-3) (x+ 1/3) = x = x and x = -1/3 are the max and min

Answer 6

x = 3 and x = -1/3......

Answer 7

f(-2) = ?
f(-1/3) = ?
f(2) = ?

Answer 8

so what do u do with the denominator

Answer 9

nothing; the top directs the =0 part; not the bottom

Answer 10

put these x values into the original equationand determine which is higher

Answer 11

awesome thanks a lot man the denominator was throwing me off

Answer 12

there is no vertical asypmtote; so you aint gotta worry about bad values :)

Answer 13

also x^2+1=0 has no real solution so this does not give you a critcal number anyways

Answer 14

right what amistre said no vertical asym

Answer 15

awesome thanks a lot guys