You visited us 1 times! Enjoying our articles? Unlock Full Access!

Cube Root of a Complex Number

If fx=2 x+1 a...

Question

If $f (x) = 2 x + 1$ and $g (x) = x^{2} + 1$ , then $g o (f o f) (2) =$

Open in App

Solution

$g o (f o f) (x) = g (f (f (x)))$
$= g (f (2 x + 1))$
$= g (2 (2 x + 1) + 1)$
$= g (4 x + 3)$
$= (4 x + 3)^{2} + 1$
$∴ (g o (f o f)) (2) = (11)^{2} + 1 = 122$

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}

f (x) = 2 x + 1

g (x) = x^{2} + 1

(g o (f o f)) (2) =

f (x) = \frac{x}{\sqrt{1 - x^{2}}}

f (x) = \frac{x}{\sqrt{1 + x^{2}}}

f (x) = \frac{3}{x - 2} - \frac{1}{x + 3}

g (x) = \frac{x^{2} - 4 x + 19}{x^{2} + x - 6}

f (x) = g (x)

x =

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

(f \circ g) (x) =

Join BYJU'S Learning Program