to find the inverse
1. replace f(x) or g(x) or f(n) with y
2. switch places of x and y
3. solve for y
4. replace y with f⁻¹(x) or g⁻¹(x) or f⁻¹(n)
1.
g(x)=3+(x-1)^3
replace
y=3+(x-1)^3
switch
x=3+(y-1)^3
solve
x=3+(y-1)^3
subtract 3
x-3=(y-1)^3
cube root both sides
\(\sqrt[3]{x-3}\)=y-1
add 1
\(\sqrt[3]{x-3}\)+1=y
switch y with g⁻¹(x)
g⁻¹(x)=\(\sqrt[3]{x-3}\)+1
4. f(x)=x^3+2
replace with y
y=x^3+2
switch x and y
x=y^3+2
solve for y
x=y^3+2
subtract 2
x-2=y^3
cube roo both sides
\(\sqrt[3]{x-2}\)=y
replace with f⁻¹(x)
f⁻¹(x)=\(\sqrt[3]{x-2}\)
5.
f(n)=5\(\sqrt{-n+1/2}\)
replace
y=5\(\sqrt{-n+1/2}\)
switch n and y
n=5\(\sqrt{-y+1/2}\)
solve for y
n=5\(\sqrt{-y+1/2}\)
divide both sides by 5
n/5=n=\(\sqrt{-y+1/2}\)
square both sides
\(\frac{n^2}{25}\)=-y+1/2
subtract 1/2 from both sides
\(\frac{n^2}{25}\)-\(\frac{1}{2}\)=-y
multiply both sides by -1
-\(\frac{n^2}{25}\)+\(\frac{1}{2}\)=y
replace with f⁻¹(n)
f⁻¹(n)=-\(\frac{n^2}{25}\)+\(\frac{1}{2}\)
1. g⁻¹(x)=\(\sqrt[3]{x-3}\)+1
4. f⁻¹(x)=\(\sqrt[3]{x-2}\)
5. f⁻¹(n)=-\(\frac{n^2}{25}\)+\(\frac{1}{2}\)
