Finding the Quotients of two Functions: f(x) = x^2 + 6, g(x) = sqrt(1 - x)

[sowieso]

Hi,
So I was wondering if someone could explain in the following example why the square root of 1-x goes on top of the fraction in the solution and why 1-x is on the bottom. The question is to find the domain.

. . .9. \(\displaystyle \, f(x)\, =\, x^2\, +\, 6,\, g(x)\, =\, \sqrt{\strut 1\, -\, x\,}\)

the answer the textbook gives is:

. . .\(\displaystyle \dfrac{(x^2\, +\, 6)\, \sqrt{\strut 1\, -\, x\,}}{1\, -\, x};\)

Thanks heaps in advance .
(and sorry if I posted in the wrong section).

 

[Deleted member 4993]

Hi,
So I was wondering if someone could explain in the following example why the square root of 1-x goes on top of the fraction in the solution and why 1-x is on the bottom. The question is to find the domain.

. . .9. \(\displaystyle \, f(x)\, =\, x^2\, +\, 6,\, g(x)\, =\, \sqrt{\strut 1\, -\, x\,}\)

the answer the textbook gives is:

. . .\(\displaystyle \dfrac{(x^2\, +\, 6)\, \sqrt{\strut 1\, -\, x\,}}{1\, -\, x};\)

Thanks heaps in advance .
(and sorry if I posted in the wrong section)..

Please post the COMPLETE problem. It should be telling you to calculate the domain of f(x)/g(x) - or some such thing!

 

[pka]

Hi,
So I was wondering if someone could explain in the following example why the square root of 1-x goes on top of the fraction in the solution and why 1-x is on the bottom. The question is to find the domain.

The answer the textbook gives is:

HINT: \(\displaystyle \dfrac{1}{\sqrt{1-x}}=\dfrac{\sqrt{1-x}}{1-x}\)

 

[Dale10101]

What are you asking?

Hi,
So I was wondering if someone could explain in the following example why the square root of 1-x goes on top of the fraction in the solution and why 1-x is on the bottom. The question is to find the domain.

. . .9. \(\displaystyle \, f(x)\, =\, x^2\, +\, 6,\, g(x)\, =\, \sqrt{\strut 1\, -\, 1\,}\)

the answer the textbook gives is:

. . .\(\displaystyle \dfrac{(x^2\, +\, 6)\, \sqrt{\strut 1\, -\, x\,}}{1\, -\, x};\)

Thanks heaps in advance .
(and sorry if I posted in the wrong section).

You seem not to be asking how to transform

\(\displaystyle \frac{{{x^2} + 6}}{{\sqrt {1 - x} }}\) to \(\displaystyle \frac{{\left( {{x^2} + 6} \right)\sqrt {1 - x} }}{{1 - x}}\)

but why one should bother.

In this case it seems a matter of protocol, perhaps because in most cases a next step is facilitated by the preferred form, or perhaps because having a preferred form makes the comparison of two expressions simpler.

Sometimes however, changing the form of an expression can change it’s implied domain so one needs to keep an eye on the original domain of the expression and it's final domain.

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from: https://www.quora.com/Mathematics/Is-sqrt-0-defined-or-undefined

…. two expressions, \(\displaystyle \sqrt x \) and \(\displaystyle \frac{x}{{\sqrt x }}\), represent the same function for x>0, but the second expression is undefined for x=0, while the first is well defined.