If f(x) = (x+1)^2 - 1 for x geq -1,then find | vedclass

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

Question

(A) Given $f(x) = (x+1)^2 - 1$ for $x \geq -1$. To find $f^{-1}(x)$,set $y = (x+1)^2 - 1$. Then $y+1 = (x+1)^2$. Since $x \geq -1$,$x+1 = \sqrt{y+1}$,

Vedclass · Accepted Answer

$\{0, -1\}$

Vedclass · Answer

(A) Given $f(x) = (x+1)^2 - 1$ for $x \geq -1$. To find $f^{-1}(x)$,set $y = (x+1)^2 - 1$. Then $y+1 = (x+1)^2$. Since $x \geq -1$,$x+1 = \sqrt{y+1}$,so $x = \sqrt{y+1} - 1$. Thus,$f^{-1}(x) = \sqrt{x+1} - 1$. The equation $f(x) = f^{-1}(x)$ is equivalent to $f(x) = x$ because $f$ is an increasing function for $x \geq -1$. So,$(x+1)^2 - 1 = x$. $x^2 + 2x + 1 - 1 = x$. $x^2 + x = 0$. $x(x+1) = 0$. This gives $x = 0$ or $x = -1$. Both values satisfy the condition $x \geq -1$. Therefore,the set is $\{0, -1\}$.

If $f(x) = (x+1)^2 - 1$ for $x \geq -1$,then find the set $\{x \mid f(x) = f^{-1}(x)\}$.

Similar Questions

If $f(x) = 3x - 5$,then ${f^{ - 1}}(x)$ is:

If $f(x) = \exp(2x^3 + 3x^2 + 6x)$ and $g(x)$ is the inverse function of $f(x)$,then the value of $g'(e^{11})$ is -

Let $f(x)=(x+1)^2-1$ for $x \geq -1$.
Statement-$1$: $S=\{x:f(x)=f^{-1}(x)\}=\{0, -1\}$
Statement-$2$: $f$ is a bijection.

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

$f(x) = \sin x + \cos x, g(x) = x^2 - 1$. Then $g(f(x))$ is invertible if:

Vedclass Test Series

Exam Paper Generator

Online Exam Module