Question #9a31c

2 Answers
Apr 8, 2017

# f^(-1) (x) = 4+-sqrt ((x + 2)/2) #

Explanation:

#f(x) = 2(x -4)^2 - 2#

let say # y =f(x)#, then #x = f^(-1) (y)#

# 2(x -4)^2 - 2 = y#

# 2(x -4)^2 = y + 2#

# (x -4)^2 = (y + 2)/2#

# (x -4) = +-sqrt ((y + 2)/2)#

# x = 4+-sqrt ((y + 2)/2) = f^(-1) (y)#

therefore,

# f^(-1) (x) = 4+-sqrt ((x + 2)/2) #

Apr 8, 2017

Technically speaking, quadratic functions are not invertible, because the ONLY point where two x values do NOT map to a single y value is the vertex. But let's proceed.

Explanation:

Given: #f(x) = 2(x-4)^2 -2#

Substitute#f^-1(x)# for every x:

#f(f^-1(x)) = 2(f^-1(x)-4)^2 -2#

Use the property #f(f^-1(x)) = x# on the left side:

#x = 2(f^-1(x)-4)^2 -2#

Subtract 2 from both sides:

#x-2 = 2(f^-1(x)-4)^2#

Divide both sides by 2:

#x/2-1 = (f^-1(x)-4)^2#

Use the square root on both sides:

#+-sqrt(x/2-1) = f^-1(x)-4#

Please observe that, if you want to restrict the inverse to returning only real values, you should restrict x to be greater than or equal to 2, at this point.

Add 4 to both sides:

#f^-1(x)= 4+-sqrt(x/2-1)#

This is two functions:

#f^-1(x)= 4+sqrt(x/2-1)# and #f^-1(x)= 4-sqrt(x/2-1)#

To assure that these are inverses of the original, you should verify that #f(f^-1(x)) = x# and #f^-1(f(x))=x#