Question

If f : [1, ∞) → [1, ∞) defined by f(x) = $2^{x(x - 1)}$, then find f⁻¹ (x).

Answer 1

Functions and Inverses

Definition of a Function: A function f from A to B is a relation such that every element in Set A has one and only one image in Set B.

Here we are given the function:

$\sf f: [1,\infty] \to [1,\infty] \\\\ f(x)=2^{x(x-1)}$

A function can have an inverse if and only if it is bijective.

We need to check bijection first.

For a function to be bijective, it must be both injective (one-one) and surjective (onto).

Domain is given as $[1,\infty ]$ and it is a valid one. Exponential Functions are valid over the entire Real Number Line.

Let us see the corresponding Range.

$\displaystyle\sf f(x)=2^{x(x-1)} = 2^{x^2-x} = \frac{2^{x^2}}{2^x}}$

Since $\sf x\geqslant 1$, $\sf 2^{x^2}$ and $\sf 2^x$ both are increasing functions.

But $\sf 2^{x^2}$ increases faster than $\sf 2^x$. So f is overall also an increasing function.

So, f is injective, since one value in range corresponds only to a single unique value in domain. In other words, every image has a unique pre-image.

Let us check for Surjectivity.

Minimum Value of x is 1.

At x=1, $\sf f(1) = 2^{1(1-1)} = 2^0 = 1$

And since f is always increasing, it will also attain all values greater than 1 for x>1.

So, Co-domain = Range. Hence, f is surjective as well.

So, we conclude that f is bijective. So inverse exists.

____________________

Finding Inverse

$\displaystyle\sf\bullet\ \textsf{Let y=f(x)}\\\\\\\implies f(x)=y=2^{x(x-1)}\\\\\\\bullet\ \textsf{Now we need x in terms of y}\\\\\\\implies x(x-1)=\log_2y \\\\\\ \implies x^2-x-\log_2y=0 \\\\\\ \textsf{This is a Quadratic Equation}\\\\\\ \textsf{Compare with \sf ax^2+bx+c=0 }\\\\\\\implies a=1,\ b=-1,\ c=-\log_2y\\\\\\ \rightarrow D=b^2-4ac=1+4\log_2y\\\\\\ \implies x=\frac{-b\pm\sqrt{D}}{2a}=\frac{1\pm\sqrt{1+4\log_2y}}{2}\\\\\\ \textsf{Rejecting the negative solution}$

$\displaystyle\sf \implies x=\frac{1+\sqrt{1+4\log_2y}}{2} \\\\\\ \bullet\ \textsf{Interchange x and y to get inverse function}\\\\\\ \implies f^{-1}: y=\frac{1+\sqrt{1+4\log_2x}}{2}\\\\\\\\ \implies \huge \boxed{\sf f^{-1}(x)=\frac{1+\sqrt{1+4\log_2x}}{2}}$