Practicing Success
Let $f:[4, ∞) → [4, ∞)$ be defined by $f(x)=5^{x(x-4)}$. Then, $f^{-1}(x)$ is |
$2-\sqrt{4+\log_5x}$ $2+\sqrt{4+\log_5x}$ $(\frac{1}{5})^{x(x-4)}$ not defined |
$2+\sqrt{4+\log_5x}$ |
The correct answer is Option (2) → $2+\sqrt{4+\log_5x}$ Clearly, $f: [4, ∞) → [4, ∞)$ is a bijection. So, it is invertible. Let $f(x) = y$. Then, $5^{x(x-4)}=y$ $⇒x^2 - 4x = \log_5 y$ $⇒x^2-4x-\log_5 y = 0$ $⇒x=\frac{4±\sqrt{16+ 4 \log_5 y}}{2}$ $⇒f^{-1}(y)=2+\sqrt{4 +\log_5 y}$ Hence, $f^{-1}(x)=2+\sqrt{4 +\log_5 x}$ We know that if g(x) is inverse of a bijection f(x), then $fog (x)=x⇒f(g(x))=x$ This relation suggests the following method for finding the inverse of a bijection. |