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 $f(x) = \frac{(\tan 1^{\circ}) x + \log_{e}(123)}{x \log_{e}(1234) - (\tan 1^{\circ})}$,$x > 0$,then the least value of $f(f(x)) + f(f(4/x))$ is $...........$.

If $f(x) = \sin^2 x$ and the composite function $g(f(x)) = |\sin x|$,then the function $g(x)$ is equal to

If $f(x) = \frac{2x+3}{3x-2}$,$x \neq \frac{2}{3}$,then the function $f \circ f$ is

Let $R$ be the set of real numbers and the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x) = x^{2} + 2x - 3$ and $g(x) = x + 1$. Then, the value of $x$ for which $f(g(x)) = g(f(x))$ is

If $f(x) = (p - x^n)^{1/n}$, $p > 0$ and $n$ is a positive integer, then $f[f(x)]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module