If the function is f(x)=frac{1}{x+2},then the | vedclass

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

Question

(A) Given the function $f(x) = \frac{1}{x+2}$.For the composite function $y = f(f(x))$,we first note that $f(x)$ is undefined at $x = -2$.Now,calculat

Vedclass · Accepted Answer

$-\frac{5}{2}$

Vedclass · Answer

(A) Given the function $f(x) = \frac{1}{x+2}$.For the composite function $y = f(f(x))$,we first note that $f(x)$ is undefined at $x = -2$.Now,calculate $f(f(x)) = f\left(\frac{1}{x+2}ight) = \frac{1}{\frac{1}{x+2} + 2}$.Simplify the denominator: $\frac{1}{x+2} + 2 = \frac{1 + 2(x+2)}{x+2} = \frac{1 + 2x + 4}{x+2} = \frac{2x + 5}{x+2}$.Thus,$f(f(x)) = \frac{x+2}{2x+5}$.The composite function is undefined when the denominator $2x+5 = 0$,which gives $x = -\frac{5}{2}$.Also,the original function $f(x)$ must be defined,so $x 
eq -2$.Therefore,the points of discontinuity are $x = -2$ and $x = -\frac{5}{2}$.Comparing with the given options,the correct point of discontinuity is $-\frac{5}{2}$.

If the function is $f(x)=\frac{1}{x+2}$,then the point of discontinuity of the composite function $y=f(f(x))$ is

Similar Questions

If $f:[0,3] \rightarrow [0,3]$ is defined by $f(x) = \begin{cases} 1+x, & 0 \leq x \leq 2 \\ 3-x, & 2 < x \leq 3 \end{cases}$,then $f(f(x))$ is:

If $g(x)=3x^{2}+2x-3,$ $f(0)=-3$ and $4g(f(x))=3x^{2}-32x+72,$ then $f(g(2))$ is equal to:

Define the functions $f, g$ and $h$ from $R$ to $R$ such that $f(x) = x^2 - 1, g(x) = \sqrt{x^2 + 1}$ and $h(x) = \begin{cases} 0, & x \leq 0 \\ x, & x \geq 0 \end{cases}$ Consider the following statements:

If $f(x)=e^x$ and $h(x)=(f \circ f)(x)$, then $\frac{h^{\prime}(x)}{h(x)}=$

If $f : R \rightarrow R$ is defined by $f(x) = x^{2} - 3x + 2$,find $f(f(x))$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module