Animated Solution for Mathematics - Functions: If the function $f:[1,\infty )\to [1,\infty )$ is defined by $f(x)=2^{x(x-1)}$, then $f^{-1}(x)$ is

The function provided is $f(x)=2^{x(x-1)}$ with the domain restricted to $[1,\infty )$. This restriction is crucial because the exponent $x^2-x$ represents a parabola with a vertex at $x=0.5$.

By restricting the domain to $x\ge 1$, we ensure the function is strictly increasing and one-to-one. This condition is necessary for the inverse function to exist.

To find the inverse, we set $y=f(x)$, which yields:

To isolate $x$, we apply the logarithm base 2 to both sides of the equation:

Since the logarithm and the exponential function are inverse operations, they simplify to:

We now treat the equation as a quadratic in terms of $x$. Rearranging into standard form $a{x}^2+bx+c=0$, we get:

Here, the coefficients are $a=1$, $b=-1$, and $c=-{\log }_{2}y$. Applying the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we substitute our values:

Simplifying the expression, we obtain:

We must choose between the positive and negative roots based on our domain constraint $x\ge 1$. If we select the negative root, the numerator becomes $1-\sqrt{1+4{\log }_{2}y}$.

Since ${\log }_{2}y\ge 0$ for $y\ge 1$, the square root term is at least 1. This would result in $x\le 0$, which violates our domain constraint.

Therefore, we must reject the negative root and retain only the positive one:

To express the inverse function $f^{-1}(x)$, we swap the variables $x$ and $y$. The final inverse function is:

Geometrically, this function represents the reflection of the original curve across the line $y=x$. You have successfully navigated the algebraic steps to determine the inverse.