A student says that the functions f(x) = 2x+2 and g(x) =2x-2 are inverse functions because their graphs are parallel. Is the student's reasoning correct? Justify your answer
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OpenStudy (anonymous):
@Nnesha
OpenStudy (anonymous):
Please help me @Nnesha
OpenStudy (anonymous):
@jim_thompson5910
OpenStudy (anonymous):
Someone please help me....
jimthompson5910 (jim_thompson5910):
what do you get if you solve y = 2x+2 for x?
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OpenStudy (anonymous):
Umh, you would get y=x-2/2
OpenStudy (anonymous):
Right?
OpenStudy (anonymous):
And then the other one would be y=x+2/2
jimthompson5910 (jim_thompson5910):
yes \[\Large y = \frac{x-2}{2}\]
that is equivalent to \[\Large y = \frac{1}{2}x - 1\]
what is the slope of the second equation I wrote?
OpenStudy (anonymous):
1/2
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OpenStudy (anonymous):
or would it be -1? Because I know that would be the y intercept right
jimthompson5910 (jim_thompson5910):
it's 1/2
OpenStudy (anonymous):
Ok
jimthompson5910 (jim_thompson5910):
so in y = 2x+2, the slope is 2
in y = (1/2)x - 1, the slope is 1/2
jimthompson5910 (jim_thompson5910):
the slopes are not equal, so the lines aren't parallel
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OpenStudy (anonymous):
And that would mean that the two functions are not inverses, right?
jimthompson5910 (jim_thompson5910):
no when we solved for x and swapped x and y, we found the inverse
OpenStudy (anonymous):
But don't inverse functions have to equal x?
jimthompson5910 (jim_thompson5910):
I'm just saying that not all inverses are parallel
OpenStudy (anonymous):
No i know that, but are they in this scenario?
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OpenStudy (anonymous):
Are the two equations inverses? I wouldn't think so because they aren't equal to x... Or am I wrong
jimthompson5910 (jim_thompson5910):
oh you mean y=2x+2 and y=2x-2?
OpenStudy (anonymous):
Yeah
jimthompson5910 (jim_thompson5910):
well you just found the inverse of y = 2x+2 was y = (1/2)x-1
so it is NOT y=2x-2
jimthompson5910 (jim_thompson5910):
y=2x+2 and y=2x-2 aren't inverses of each other
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OpenStudy (anonymous):
Oh ok. Yeah, that's what I thought, but I just wanted to make sure. Do you think you could help me with another problem?
OpenStudy (anonymous):
Another problem that also has to do with inverse functions...?
jimthompson5910 (jim_thompson5910):
go ahead
OpenStudy (anonymous):
Find the inverse of each function. Then use the definition of inverse functions to verify that the two functions are inverses.
1. f(x) = -3x+3
OpenStudy (anonymous):
I know that the inverse for this one is f -1(x)=3-x/3
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OpenStudy (anonymous):
The other function is g(x)=0.25x+.6
jimthompson5910 (jim_thompson5910):
use parenthesis and say (3-x)/3
jimthompson5910 (jim_thompson5910):
keep in mind that 3-x/3 without parenthesis means \(\LARGE 3 - \frac{x}{3}\)
OpenStudy (anonymous):
Ok. And the inverse for the second one is f-1(x)=-4(.6-x) right?
OpenStudy (anonymous):
Would I have to plug them into each other to see if they are inverses following the definition of inverse functions?
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jimthompson5910 (jim_thompson5910):
y=0.25x+0.6 has the inverse y = 4(x-0.6) or y = 4x-2.4
jimthompson5910 (jim_thompson5910):
yeah you need to confirm that f(g(x)) = x and g(f(x)) = x
OpenStudy (anonymous):
ok, so what would I do next?
OpenStudy (anonymous):
Ok. and they are not inverses right?
jimthompson5910 (jim_thompson5910):
which 2?
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OpenStudy (anonymous):
there's more than two?
jimthompson5910 (jim_thompson5910):
I'm lost about what you're asking
OpenStudy (anonymous):
You asked me which two
OpenStudy (anonymous):
What does that mean?
OpenStudy (anonymous):
I apologize. I just don't understand this
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jimthompson5910 (jim_thompson5910):
so you're given 2 functions, right? which 2 functions are you given again?
OpenStudy (anonymous):
I don't know
OpenStudy (anonymous):
Would it be the two inverses of the functions?
jimthompson5910 (jim_thompson5910):
f(x) = -3x+3 and g(x)=0.25x+.6 right?
OpenStudy (anonymous):
Yes
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jimthompson5910 (jim_thompson5910):
so if you can show that f(g(x)) = x and g(f(x)) = x are both true, then you have proven they are inverses of each other
OpenStudy (anonymous):
ok. Thank you for your help.. I'll post another question if I need help.