How do you find $(\frac{g}{h}) (0)$ $(g/h)(0)$ given $f (x) = x^{2} - 1 and$ $f(x)=x^2-1 and$ g(x)=2x-3 $and$ $and$ h(x)=1-4x#?

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How do you find $(\frac{g}{h}) (0)$ $(g/h)(0)$ given $f (x) = x^{2} - 1 and$ $f(x)=x^2-1 and$ g(x)=2x-3 $and$ $and$ h(x)=1-4x#?

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Answer 1 · 2017-02-15T16:38:29

$(\frac{g}{h}) (0) = - 3$ $(g/h)(0) = -3$

Explanation:

$(\frac{g}{h}) (x) = \frac{2 x - 3}{1 - 4 x}; (\frac{g}{h}) (0) = \frac{2 (0) - 3}{1 - 4 (0)} = - 3$ $(g/h)(x) = (2x-3)/(1-4x); (g/h)(0) = (2(0)-3)/(1-4(0)) = -3$

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How do you find $(\frac{g}{h}) (0)$ $(g/h)(0)$ given $f (x) = x^{2} - 1 and$ $f(x)=x^2-1 and$ g(x)=2x-3 $and$ $and$ h(x)=1-4x#?

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How do you find $(\frac{g}{h}) (0)$ $(g/h)(0)$ given $f (x) = x^{2} - 1 and$ $f(x)=x^2-1 and$ g(x)=2x-3 $and$ $and$ h(x)=1-4x#?