Given f(x)=4x-3, g(x)=1/x and h(x)= x^2-x, how do you find h(1/x)?

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Given f(x)=4x-3, g(x)=1/x and h(x)= x^2-x, how do you find h(1/x)?

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Answer 1 · 2017-09-13T17:44:56

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

Explanation:

$to evaluate h (\frac{1}{x}) substitute x = \frac{1}{x} in h (x)$ $"to evaluate "h(1/x)" substitute "x=1/x" in "h(x)$

$\Rightarrow h (\frac{1}{x}) = {(\frac{1}{x})}^{2} - \frac{1}{x}$ $rArrh(1/x)=(1/x)^2-1/x$

$\Rightarrow h (\frac{1}{x}) = \frac{1}{x^{2}} - \frac{1}{x}$ $color(white)(rArrh(1/x))=1/x^2-1/x$

$we can express this as a single fraction$ $"we can express this as a single fraction"$

$multiplt the numerator/denominator of \frac{1}{x} by x$ $"multiplt the numerator/denominator of "1/x" by "x$

$\Rightarrow \frac{1}{x^{2}} - \frac{x}{x^{2}}$ $rArr1/x^2-x/x^2$

$denominators are common so subtract the numerators$ $"denominators are common so subtract the numerators"$
$leaving the denominator as it is.$ $"leaving the denominator as it is."$

$\Rightarrow h (\frac{1}{x}) = \frac{1}{x^{2}} - \frac{x}{x^{2}} = \frac{1 - x}{x^{2}}$ $rArrh(1/x)=1/x^2-x/x^2=(1-x)/x^2$

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Given f(x)=4x-3, g(x)=1/x and h(x)= x^2-x, how do you find h(1/x)?

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