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jimthompson5910 (jim_thompson5910):
f(x) = 5x + 3
y = 5x + 3
x = 5y + 3
Now solve for y
OpenStudy (anonymous):
I dont understand
jimthompson5910 (jim_thompson5910):
What do you get when you solve for y
jimthompson5910 (jim_thompson5910):
I started off by replacing f(x) with y. Then I swapped x and y. The goal now is to solve for y.
OpenStudy (anonymous):
plug in what you got for x
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jimthompson5910 (jim_thompson5910):
what do you mean
OpenStudy (anonymous):
plug in what I got for x into the y= equation
jimthompson5910 (jim_thompson5910):
no, just tell me what you get when you solve for y
OpenStudy (anonymous):
5
jimthompson5910 (jim_thompson5910):
solve
x = 5y + 3
for y
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OpenStudy (anonymous):
y = 3/-5
jimthompson5910 (jim_thompson5910):
\[\Large x = 5y + 3\]
\[\Large x-3 = 5y\]
\[\Large \frac{x-3}{5} = y\]
\[\Large y=\frac{x-3}{5} \]
So the inverse function is
\[\Large f^{-1}(x)=\frac{x-3}{5} \]
OpenStudy (anonymous):
thank you but where did the f^-1 come from
jimthompson5910 (jim_thompson5910):
that stands for the inverse function
jimthompson5910 (jim_thompson5910):
once you've solved for y, that's effectively finding the inverse
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