Let $f (x) = {\begin{matrix} 1 + x & 0 \leq x \leq 2 3 - x & 2 \leq x \leq 3 \end{matrix}$ . Determine the composite function g(x)=f(x) & hence find the point of discontinuity of g, if any.

Open in App

Solution

$f (x) = {\begin{matrix} 1 + x & 0 \leq x \leq 2 3 - x & 2 \leq x \leq 3 \end{matrix}$

$g (x) = f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 2 + x & 0 \leq x \leq 1 2 - x & 1 \leq x \leq 2 4 - x & 2 \leq x \leq 3 \end{matrix}$

$1 (1 -) = 3 g (1 +) = 1$

$g (2 -) = 0 g (2 +) = 2$

Discontinuous at $x = 1$ & $x = 2$

Suggest Corrections

Similar questions

f (x) = 1 + x f o r 0 \leq x \leq 2

= 3 - x f o r 2 < x \leq 3.

g (x) = f (f (x))

How many of the following pairs of function are identcal

(a) $f (x) = \frac{x^{2}}{x}, g (x) = x$

(b) $f (x) = \frac{x^{2} - 1}{x - 1}, g (x) = x + 1$

(d) $f (x) = \frac{x}{x^{2}}, g (x) = \frac{1}{x}$

f (x) = \frac{1}{x - 1}

y = f {f (x)}

f (x) = \frac{1}{x - 1}

y = [f {f (x)}]

f (x) = \frac{1}{x - 1} .

f [f {f (x)}] .

Join BYJU'S Learning Program