Inverse functions can be solved in the following way (example)

If we have a function \(f(x) = y = mx + c\)
Then \(f^{-1}(x)\) can be solved by interchanging x and y values i.e. \(f^{-1}(x) = x = my + c\)

NOTE: Do not confuse inverse of function with the reciprocal of function
\(\frac{1}{f(x)}\) is the reciprocal not the inverse!

Also, Inverse is written as \(f^{-1}(x)\) whereas, Reciprocal would be \([f(x)]^{-1}\)

Now we are given,
\(f(x) = 3x - 15\)
So, \(f^{-1}(-3) = 3y - 15\)
\(-3 = 3y - 15\)
\(y = 4\)

\(g(x) = (x+3)^2 + 3\)
\(g^{-1}(4) = (y+3)^2 + 3\)
\(4 = (y+3)^2 + 3\)
\(1 = (y+3)^2\)
\(y = -2\) or \(-4\)

Col. A: \(4\)
Col. B: \(-2\) or \(-4\)

Hence, option A

Let's learn to find Inverse function \(f^{-1}(x)\) first

To find \(f^{-1}(x)\) we need to write x in terms of f(x)
\(f(x)=3x-15\)
=> 3x = f(x) + 15
=> x = \(\frac{f(x)}{3} + 5\)
Now replace x with \(f^{-1}(x)\) we get
\(f^{-1}(x)\) = \(\frac{f(f^{-1}(x))}{3}\) + 5
=> \(f^{-1}(x)\) = \(\frac{x}{3}\) + 5

Let's find \(g^{-1}(x)\) similarly
\( g(x)=(x+3)^2+3\)
\( (x+3)^2 = g(x) - 3\)
Now there will be two cases
Case 1:
x+3 = +\(\sqrt{g(x) - 3}\)
=> x = +\(\sqrt{g(x) - 3}\) - 3
Replace x with \(g^{-1}(x)\) we get
\(g^{-1}(x)\) = \(\sqrt{g(g^{-1}(x)) - 3}\) - 3
=> \(g^{-1}(x)\) = \(\sqrt{x - 3}\) - 3

Case 2:
x+3 = -\(\sqrt{g(x) - 3}\)
=> x = -\(\sqrt{g(x) - 3}\) - 3
Replace x with \(g^{-1}(x)\) we get
\(g^{-1}(x)\) = -\(\sqrt{g(g^{-1}(x)) - 3}\) - 3
=> \(g^{-1}(x)\) = -\(\sqrt{x - 3}\) - 3


Quantity A: \(f^{-1}(-3)\)
=> \(f^{-1}(x)\) = \(\frac{x}{3}\) + 5
=> \(f^{-1}(-3)\) = \(\frac{-3}{3}\) + 5 = -1 + 5 = 4

Quantity B: \( g^{-1}(4)\)
Case 1
=> \(g^{-1}(x)\) = \(\sqrt{x - 3}\) - 3
=> \(g^{-1}(4)\) = \(\sqrt{4 - 3}\) - 3 = 1-3 = -2

Case 2
=> \(g^{-1}(x)\) = -\(\sqrt{x - 3}\) - 3
=> \(g^{-1}(4)\) = -\(\sqrt{4 - 3}\) - 3 = -1-3 = -4


Clearly, Quantity A(4) > Quantity B(-2 or -4)

So, Answer will be A
Hope it helps!

Watch the following video to MASTER Functions

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