If gx - 1 - x and hx - xx-1 then ghxhgx is

G(h(x))h(g(x)) = 1 h(x)g(x)g(x) 1=1 ( xx 1 )(1 x)(1 x) 1= x(1 x)^2 = ( xx 1 )1 x=h(x)g(x) ...

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Review of Functions

If gx = 1 - x...

Question

If $g (x) = 1 - x$ and $h (x) = \frac{x}{x - 1}$ then $\frac{g (h (x))}{h (g (x))}$ is

\frac{h (x)}{g (x)}

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\frac{- 1}{x}

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\frac{g (x)}{h (x)}

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\frac{x}{(1 - x)^{2}}

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Solution

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Similar questions

g (x) = 1 - x

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

\frac{g (h (x))}{h (g (x))}

If $f (x) = x^{2} + 1$ and $g (x) = x^{2} - 1$ , then what is the value of f[g{h(x)}]?

Given, $h (x) = \frac{1}{55} 5 \sum x = 1 [f (x) + g (x)]$

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)}

h (x)

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

g (x) = x - \frac{1}{x}, x \in R - - 1, 0, 1

h (x) = \frac{f (x)}{g (x)}

h (x)

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

g (x) = x - \frac{1}{x},

x \in R - {- 1, 0, 1} .

h (x) = \frac{f (x)}{g (x)},

h (x)

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