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Zeroes of a Polynomial

If fx = ...

Question

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

A

x

- 2

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B

2 -

x

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C

x

+ 2

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D

-

(x + 2)

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Solution

The correct option is C $x$ + 2
Given, $f (x) = x + 2$ .....(1) and $g (x) = x$ .....(2).
Now,
$(g o f) (x)$
$= g [f (x)]$
$= f (x)$ { Using (2)]
$= x + 2$ . [ Using (1)]

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Similar questions

f (x) = x + 2; g (x) = x; h (x) = x^{2}

h o (g o f) (x) =

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

g (x) = \frac{x - 3}{x + 4}

h (x) = - \frac{2 (2 x + 1)}{x^{2} + x - 12}

lim x \to 3 [f (x) + g (x) + h (x)]

f (x) = \frac{2}{x - 3}, g (x) = \frac{x - 3}{x + 4}

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lim x \to 3 [f (x) + g (x) + h (x)]

f (x) = x^{2} + \frac{1}{x^{2}} and g (x) = x - \frac{1}{x}, x ϵ R - {- 1, 0, 1}, and h (x) = \frac{f (x)}{g (x)}

then the local minimum value of

h (x)

f (x) = [\frac{x}{[x]}], g (x) = [x [\frac{1}{x}]]

h (x) = [\frac{[x]}{x}],

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