If f:[1, +infty) to [2, +infty) is given by f | vedclass

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

Question

(A) Given $y = f(x) = x + \frac{1}{x}$ where $x \ge 1$ and $y \ge 2$. Multiplying by $x$,we get $yx = x^2 + 1$,which rearranges to the quadratic equat

Vedclass · Accepted Answer

$\frac{x + \sqrt{x^2 - 4}}{2}$

Vedclass · Answer

(A) Given $y = f(x) = x + \frac{1}{x}$ where $x \ge 1$ and $y \ge 2$. Multiplying by $x$,we get $yx = x^2 + 1$,which rearranges to the quadratic equation $x^2 - yx + 1 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{y \pm \sqrt{y^2 - 4}}{2}$. Since the domain of $f$ is $[1, +\infty)$,we must have $x \ge 1$. If we take the negative sign,$x = \frac{y - \sqrt{y^2 - 4}}{2}$. For $y \ge 2$,this value is $\le 1$. Specifically,as $y 	o \infty$,this expression approaches $0$. Thus,to satisfy $x \ge 1$,we must take the positive sign: $x = \frac{y + \sqrt{y^2 - 4}}{2}$. Replacing $y$ with $x$,we obtain ${f^{-1}}(x) = \frac{x + \sqrt{x^2 - 4}}{2}$.

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

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

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.

Consider $f: \{1, 2, 3\} \rightarrow \{a, b, c\}$ given by $f(1) = a, f(2) = b$ and $f(3) = c$. Find $f^{-1}$ and show that $(f^{-1})^{-1} = f$.

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?

If a function $f:(-1,1) \rightarrow B(\subseteq R)$ is defined as $f(x)=x+x^2+x^3+\ldots \infty$, then in order to have the inverse function of $f$, $B$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module