The inverse of the function f(x) = frac{10^x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.To find the inverse,we solve for $x$ in terms of $y$.Multiply the numerator and denominator by $10

Vedclass · Accepted Answer

$\frac{1}{2} \log_{10} \left( \frac{1+x}{1-x} \right)$

Vedclass · Answer

(A) Let $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.To find the inverse,we solve for $x$ in terms of $y$.Multiply the numerator and denominator by $10^x$:$y = \frac{10^{2x} - 1}{10^{2x} + 1}$.$y(10^{2x} + 1) = 10^{2x} - 1$.$y \cdot 10^{2x} + y = 10^{2x} - 1$.$1 + y = 10^{2x}(1 - y)$.$10^{2x} = \frac{1+y}{1-y}$.Taking $\log_{10}$ on both sides:$2x = \log_{10} \left( \frac{1+y}{1-y} \right)$.$x = \frac{1}{2} \log_{10} \left( \frac{1+y}{1-y} \right)$.Replacing $y$ with $x$ to get the inverse function:$f^{-1}(x) = \frac{1}{2} \log_{10} \left( \frac{1+x}{1-x} \right)$.

The inverse of the function $f(x) = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$ is

Similar Questions

If $f: R - \{\frac{3}{5}\} \rightarrow R - \{\frac{3}{5}\}$ is defined by $f(x) = \frac{3x+1}{5x-3}$,then which of the following is true?

The inverse of the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} + 1$ is $x =$

Let $x \neq 0$ and $|x| < \frac{1}{2}$. If $f(x) = 1 + 2x + 4x^2 + 8x^3 + \ldots$,then $f^{-1}(x) =$

Let $f: R - \{\frac{\alpha}{6}\} \rightarrow R$ be defined by $f(x) = \frac{5x + 3}{6x - \alpha}$. Then the value of $\alpha$ for which $(f \circ f)(x) = x$,for all $x \in R - \{\frac{\alpha}{6}\}$,is:

If the function $f(x) = x^5 + e^{x/5}$ and $g(x) = f^{-1}(x)$,then the value of $\frac{1}{g'(1 + e^{1/5})}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module