$If f (x) = x - 1, g (x) = x + 1, is f o g = g o f$

Open in App

Solution

$Given, f (x) = x - 1, g (x) = x + 1$
$f o g = f (g (x)) = f (x + 1) = x + 1 - 1 = x$

$g o f = g (f (x)) = g (x - 1) = f (x - 1) = x - 1 + 1 = x$ (1 mark)

$∴ f o g = g o f$ (1 mark)

Suggest Corrections

Similar questions

f (x) = e^{x}, g (x) = \log_{e} x

f (x) = x^{2}, g (x) = \cos x

f (x) = | x |, g (x) = \sin x

f (x) = x + 1, g (x) = e^{x}

f (x) = \sin^{- 1} x, g (x) = x^{2}

f (x) = x + 1, g (x) = \sin x

f (x) = x + 1, g (x) = 2 x + 3

f (x) = c, c \in R, g (x) = \sin x^{2}

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

If f (x) = 2 x - 1, g (x) = \frac{x + 1}{2}, then f o g - g o f =

f (x) = {\begin{matrix} 1 - | x |, | x | \leq 1 | x | - 1, | x | > 1 \end{matrix}

g (x) = f (x - 1) + f (x + 1)

5 \int - 3 g (x) d x

Join BYJU'S Learning Program