Let f(x)= [tex] \frac{x {}^{2} }{x {}^{2 } -

Ответ:

Rogeartest4

08.01.2022

Mathematics

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

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

$x1\ \land x\neq 2$

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Ответ:

xaviercarrasco04

18.02.2021

Mathematics

x=14.0°

Step-by-step explanation:

x= angle of slope

$tanx=\frac{12.0ft}{98.0ft} =0.2500$

$x=tan^{-1} (0.2500)$

x=14.0

so, the angle of slope is 14.0°

╭━━━━━━━╮

hope it helps...

have a great day

╰━━━━━━━╯

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

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.

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Let f(x)= and g(x)= a) find the domains of f (x) and g(x).b) find f (g(x)) and describe its domain.

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

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.