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 $[\cdot]$ denotes the greatest integer function and if $f:(5,10) \rightarrow(7,12)$ is a function defined by $f(x)=x+2\left[\frac{x}{5}\right]$,then

Consider $f: R_{+} \rightarrow [4, \infty)$ given by $f(x) = x^{2} + 4$. Show that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1}(y) = \sqrt{y - 4}$,where $R_{+}$ is the set of all non-negative real numbers.

If $f:[1, \infty) \rightarrow [2, \infty)$ is given by $f(x) = x + \frac{1}{x}$,then $f^{-1}(x)$ equals

$f: R \rightarrow R, f(x) = 3x + 2$ and $g: R \rightarrow R, g(x) = 6x + 5$. Find the value of $(g \circ f^{-1})(10)$.

If $f:[1, \infty) \rightarrow[5, \infty)$ is given by $f(x)=3x+\frac{2}{x}$, then $f^{-1}(x)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module