If f:[1, infty) rightarrow [0, infty) is give | vedclass

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

Question

(C) Given $f(x) = x - \frac{1}{x}$. Let $y = f(x)$, so $y = x - \frac{1}{x}$.To find $f^{-1}(x)$, we solve for $x$ in terms of $y$:$y = \frac{x^2 - 1}

Vedclass · Accepted Answer

$\frac{1}{2}\left[x+\sqrt{x^2+4}\right]$

Vedclass · Answer

(C) Given $f(x) = x - \frac{1}{x}$. Let $y = f(x)$, so $y = x - \frac{1}{x}$.To find $f^{-1}(x)$, we solve for $x$ in terms of $y$:$y = \frac{x^2 - 1}{x} \Rightarrow x^2 - yx - 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{y \pm \sqrt{y^2 - 4(1)(-1)}}{2(1)} = \frac{y \pm \sqrt{y^2 + 4}}{2}$.Since the domain of $f$ is $[1, \infty)$, $x$ must be positive, so we take the positive root:$x = \frac{y + \sqrt{y^2 + 4}}{2}$.Replacing $y$ with $x$, we get $f^{-1}(x) = \frac{x + \sqrt{x^2 + 4}}{2}$.

If $f:[1, \infty) \rightarrow [0, \infty)$ is given by $f(x) = x - \frac{1}{x}$, then $f^{-1}(x) =$

Similar Questions

Let $f: N \to Y$ be a function defined as $f(x) = 4x + 3$,where $Y = \{y \in N : y = 4x + 3, x \in N\}$. Show that $f$ is invertible and find its inverse.

If $f:[1, \infty) \rightarrow [1, \infty)$ is defined by $f(x) = \frac{1+\sqrt{1+4 \log_2 x}}{2}$, then $f^{-1}(3) =$

If $f(x) = x^2 - x + 5, x > \frac{1}{2},$ and $g(x)$ is its inverse function,then $g'(7)$ equals

If $f(x) = \frac{3x+2}{5x-3}$,where $x \in R - \{\frac{3}{5}\}$,then:

Let $g(x)$ be the inverse of the function $f(x)$ and $f'(x) = \frac{1}{1 + x^3}$. Then $g'(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module