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Jlenok [28]
3 years ago
7

Let f(x)=

id="TexFormula1" title=" \frac{x {}^{2} }{x {}^{2 } - 1 } " alt=" \frac{x {}^{2} }{x {}^{2 } - 1 } " align="absmiddle" class="latex-formula">
and g(x)= \frac{1}{ \sqrt{x - 1} }

a) Find the domains of f (x) and g(x).
b) Find f (g(x)) and describe its domain.​
Mathematics
1 answer:
S_A_V [24]3 years ago
8 0

Domain:

f(x) has a denominator, which can't be zero. So, its domain is given by

x^2-1\neq 0 \iff x^2\neq 1 \iff x\neq \pm 1

g(x) has a denominator as well. Moreover, it has a root. So, the content of the root can't be negative:

x-1\geq 0 \iff x \geq 1

And the denominator can't be zero:

\sqrt{x-1}\neq 0 \iff x-1 \neq 0 \iff x \neq 1

So, the domain is x>1

Composition:

We have

f(g(x))=\dfrac{g^2(x)}{g^2(x)-1} = \dfrac{\dfrac{1}{x-1}}{\dfrac{1}{x-1}-1} = \dfrac{\dfrac{1}{x-1}}{\dfrac{1-x+1}{x-1}} = \dfrac{\dfrac{1}{x-1}}{\dfrac{2-x}{x-1}}=\dfrac{1}{2-x}

The domain of this function is

2-x\neq 0 \iff x\neq 2

But we also have to remember about the domain of g(x): if g(x) is undefined, we can't compute f(g(x))!

So, the domain of f(g(x)) is

x>1\ \land x\neq 2

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The weight of the water in the pool is approximately 60,000 lb·f

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