Question

Find the inverse of each function
1) g (x)= 3+(x-1)^3
4) f (x)=x^3+2
5) f (n)= 5 square root of (-n+1/2)

Answer 1

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)=-$\frac{n^2}{25}$+$\frac{1}{2}$