If f:[1, infty) rightarrow [2, infty) is give | vedclass

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

Question

(A) Given $f(x) = x + \frac{1}{x}$.Let $y = f(x) = x + \frac{1}{x}$.Multiplying by $x$,we get $xy = x^2 + 1$,which implies $x^2 - xy + 1 = 0$.Using th

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $f(x) = x + \frac{1}{x}$.Let $y = f(x) = x + \frac{1}{x}$.Multiplying by $x$,we get $xy = x^2 + 1$,which implies $x^2 - xy + 1 = 0$.Using the quadratic formula to solve for $x$:$x = \frac{y \pm \sqrt{y^2 - 4}}{2}$.Since the domain is $x \in [1, \infty)$,we must have $x \ge 1$.If we take $x = \frac{y - \sqrt{y^2 - 4}}{2}$,then for $y=2$,$x=1$,but for $y > 2$,this value would be less than $1$.Thus,we choose the positive root: $x = \frac{y + \sqrt{y^2 - 4}}{2}$.Therefore,$f^{-1}(x) = \frac{x + \sqrt{x^2 - 4}}{2}$.

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

Similar Questions

If $g(x)$ is the inverse function of $f(x)$ and $f^{\prime}(x) = \frac{1}{1+x^4}$,then $g^{\prime}(x)$ is

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

If ${e^x} = y + \sqrt {1 + {y^2}} $,then $y =$

Inverse of the function $y = 2x - 3$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module