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 $f: R - \{3\} \rightarrow R - \{1\}$ be defined by $f(x) = \frac{x-2}{x-3}$. Let $g: R \rightarrow R$ be given as $g(x) = 2x - 3$. Then,the sum of all the values of $x$ for which $f^{-1}(x) + g^{-1}(x) = \frac{13}{2}$ is equal to ...... .

Let $f$ and $g$ be two differentiable functions satisfying $g^{\prime}(5)=\frac{3}{4}$, $g(5)=6$ and $g=f^{-1}$. Then $f^{\prime}(6)$ is equal to

Let $f(x) = \frac{x^2 - x}{x^2 + 2x}$,$x \ne 0, -2$. Then $\frac{d}{dx}[f^{-1}(x)]$ (wherever it is defined) is equal to

If $f$ is the inverse function of $g$ and $g^{\prime}(x)=\frac{1}{1+x^n}$, then the value of $f^{\prime}(x)$ is

If $f(x) = \frac{x}{1 + x}$,then ${f^{-1}}(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module