Let f(x) = (x + 1)^2 - 1 for x ge -1. Then th | vedclass

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

Question

(D) Given $f(x) = (x + 1)^2 - 1$ for $x \ge -1$.To find the set $S = \{ x : f(x) = f^{-1}(x) \}$,we note that the solutions to $f(x) = f^{-1}(x)$ are

Vedclass · Accepted Answer

$\{0, -1, \frac{-3 + i\sqrt{3}}{2}, \frac{-3 - i\sqrt{3}}{2}\}$

Vedclass · Answer

(D) Given $f(x) = (x + 1)^2 - 1$ for $x \ge -1$.To find the set $S = \{ x : f(x) = f^{-1}(x) \}$,we note that the solutions to $f(x) = f^{-1}(x)$ are the same as the solutions to $f(x) = x$ and the points where the graph of $f(x)$ intersects its inverse.Setting $f(x) = x$,we get $(x + 1)^2 - 1 = x$.$x^2 + 2x + 1 - 1 = x \Rightarrow x^2 + x = 0 \Rightarrow x(x + 1) = 0$.Thus,$x = 0$ and $x = -1$ are solutions.However,the equation $f(x) = f^{-1}(x)$ is equivalent to $f(f(x)) = x$ for $x$ in the domain.$( (x + 1)^2 - 1 + 1 )^2 - 1 = x \Rightarrow ((x + 1)^2)^2 - 1 = x$.$(x + 1)^4 - 1 = x \Rightarrow (x + 1)^4 - (x + 1) = 0$.$(x + 1) [ (x + 1)^3 - 1 ] = 0$.This gives $x + 1 = 0 \Rightarrow x = -1$ or $(x + 1)^3 = 1$.The roots of $(x + 1)^3 = 1$ are $x + 1 = 1, \omega, \omega^2$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$.$x + 1 = 1 \Rightarrow x = 0$.$x + 1 = \frac{-1 + i\sqrt{3}}{2} \Rightarrow x = \frac{-3 + i\sqrt{3}}{2}$.$x + 1 = \frac{-1 - i\sqrt{3}}{2} \Rightarrow x = \frac{-3 - i\sqrt{3}}{2}$.Thus,the set $S = \{ 0, -1, \frac{-3 + i\sqrt{3}}{2}, \frac{-3 - i\sqrt{3}}{2} \}$.

Let $f(x) = (x + 1)^2 - 1$ for $x \ge -1$. Then the set $S = \{ x : f(x) = f^{-1}(x) \}$ is

Similar Questions

Let $f(x) = (x + 2)^2 - 2, x \geq - 2$. Then $f^{-1}(x) =$

Let $f(x) > 0$ for all $x$ and $f^{\prime}(x)$ exists for all $x$. If $f$ is the inverse function of $h$ and $h^{\prime}(x) = \frac{1}{1 + \log x}$, then $f^{\prime}(x)$ will be

Which of the following functions cannot have their inverse defined? (where $[.] \to$ greatest integer function)

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

If the function $f(x) = -4e^{\left(\frac{1-x}{2}\right)} + 1 + x + \frac{x^2}{2} + \frac{x^3}{3}$ and $g(x) = f^{-1}(x)$,then the value of $g'(-\frac{7}{6})$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module