when are they equal?
(x+2)/(x+4) = (x+2)/(x-2)
(x+2)(x-2) = (x+2)(x+4)
(x-2) =(x+4)
never
when does the denominator of f(x) = 0
when x = -4
if x < -4, like -5 , how do f and g compare?
f = -3/-1 = 3 and g = -3/-7 = 3/7
so when x<-4 f >g
now how about x between -4 and +2 when g blows up?
say x = 0
f = 2/4 and g = -1
so between x = -4 and +2, f>g
now what if x is > +2, like +3?
f = 5/7 and g = 5/1 Oh, now g>x
so I think f>g to the left of x = +2
Given f(x) = (x+2)/(x+4) and g(x) = (x+2)/(x-2) find when f(x)>=g(x)
I've plugged the reciprocals into Desmos, but I'm having difficulty
I'm also unsure of what to do with the interval chart.
2 answers
(x+2)/(x+4) ≥ (x+2)/(x-2)
clearly x ≠ -4,2
We could look at the graph of f(x) and g(x) as a start
https://www.wolframalpha.com/input/?i=graph+y+%3D+%28x%2B2%29%2F%28x%2B4%29%2C+y+%3D+%28x%2B2%29%2F%28x-2%29
Wolfram graphed f(x) in blue and g(x) in red
notice where the blue curve is above the red curve, keep in mind that there are vertical asymptotes
at x = -4 and x = 2
from the graph it is clear that f(x) > g(x) for x < -4, then again for -2 < x < 2
let's see if if can find that algebraically:
(x+2)(x-2) = (x+4)(x+2)
x^2 - 4 = x^2 + 6x + 8
6x = -12
x = -2 , which is the x of our intersection
so we have 3 critical values: x = -4, x = -2 and x = 2
which means we have to investigate:
x < -4, -4 < x < -2, -2 < x < 2, and x > 2
test for x<-4 , let's pick -5
(x+2)/(x+4) ≥ (x+2)/(x-2)
LS = -3/-1 = 3
RS = -3/-3 = 1, and LS ≥ RS, so x < -4 works
test for -4 < x < -2 , let's pick x = -3
LS = -1/1 = -1
RS = -1/-5 = 1/5 , so not true, and -4<x<-2 is out
test for -2<x<2, let x = 0
LS = 2/4 = 1/2
RS = 2/-2 = -1, the statement is true, so -2 < x < 2 works
let's test for x > 2, let's pick x = 5
LS = 7/9
RS = 7/3 and since 7/9 < 7/3, x>2 is out
so we have :
x < -4 OR -2 < x < 2 , as seen in the graph
clearly x ≠ -4,2
We could look at the graph of f(x) and g(x) as a start
https://www.wolframalpha.com/input/?i=graph+y+%3D+%28x%2B2%29%2F%28x%2B4%29%2C+y+%3D+%28x%2B2%29%2F%28x-2%29
Wolfram graphed f(x) in blue and g(x) in red
notice where the blue curve is above the red curve, keep in mind that there are vertical asymptotes
at x = -4 and x = 2
from the graph it is clear that f(x) > g(x) for x < -4, then again for -2 < x < 2
let's see if if can find that algebraically:
(x+2)(x-2) = (x+4)(x+2)
x^2 - 4 = x^2 + 6x + 8
6x = -12
x = -2 , which is the x of our intersection
so we have 3 critical values: x = -4, x = -2 and x = 2
which means we have to investigate:
x < -4, -4 < x < -2, -2 < x < 2, and x > 2
test for x<-4 , let's pick -5
(x+2)/(x+4) ≥ (x+2)/(x-2)
LS = -3/-1 = 3
RS = -3/-3 = 1, and LS ≥ RS, so x < -4 works
test for -4 < x < -2 , let's pick x = -3
LS = -1/1 = -1
RS = -1/-5 = 1/5 , so not true, and -4<x<-2 is out
test for -2<x<2, let x = 0
LS = 2/4 = 1/2
RS = 2/-2 = -1, the statement is true, so -2 < x < 2 works
let's test for x > 2, let's pick x = 5
LS = 7/9
RS = 7/3 and since 7/9 < 7/3, x>2 is out
so we have :
x < -4 OR -2 < x < 2 , as seen in the graph