Asked by Muneer
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.
I've plugged the reciprocals into Desmos, but I'm having difficulty
I'm also unsure of what to do with the interval chart.
Answers
Answered by
Damon
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
(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
Answered by
Reiny
(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 <b> x < -4</b> 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 <b> x < -4</b> 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
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