SOLUTION: Please help me solve this: {{{ find f^-1 }}} {{{ f(x)=(x+2)^2 }}} x is greater than or equal to -2

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Question 930211: Please help me solve this: +find+f%5E-1++ +f%28x%29=%28x%2B2%29%5E2++ x is greater than or equal to -2


Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me solve this: +find+f%5E-1++ +f%28x%29=%28x%2B2%29%5E2++ x is greater than or equal to -2
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Note: y = (x+2)^2 is a parabola opening upward from vertex = (-2,0)
That function does not have an inverse as each x-value would have
2 corresponding y-values.
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But, if x >= -2, the function DOES have an inverse.
1st: Interchange x and y to get:
x = (y+2)^2
2nd: Solve for "y"::
y+2 = sqrt(x)
f^=1 = sqrt(x)-2 and y >= -2
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Cheers,
Stan H.
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Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
+f%28x%29=%28x%2B2%29%5E2++
to +find+f%5E-1++ first recall that +f%28x%29=y++

y=%28x%2B2%29%5E2++...swap x and y

x=%28y%2B2%29%5E2++...solve for y

sqrt%28x%29=sqrt%28%28y%2B2%29%5E2%29++

sqrt%28x%29=y%2B2++

y=%28-2+%2B-+sqrt%28x+%29%29

so, +find+f%5E-1+=%28-2+%2B-+sqrt%28x+%29%29+