functions: find g(f(x)) for f = 3x + 1, g = (x^2 + 5x)^(1/2)

[Jodene222]

If f(x) = 3x + 1, and g(x) = (x^2 + 5x)^-1/2, find g (f(x))

my answer is:

1
_________________
3(3x^2 + 7x + 2) (all in the square root sign, sorry, not sure how to type this)


I originally had the square root of 9x^2 + 21X + 6 and I reduced it by taking out 3.
please let me know if I am on the right track.
THANKS!

 

[pka]

Re: functions question

\(\displaystyle \begin{array}{l} g(x) = \left( {x^2 + 5x} \right)^{ - \frac{1}{2}} = \frac{1}{{\sqrt {x^2 + 5x} }} \\ g(f(x)) = \frac{1}{{\sqrt {\left( {f(x)} \right)^2 + 5\left( {f(x)} \right)} }} \\ \end{array}\)

 

[Jodene222]

IS MY ANSWER ALSO CORRECT?

 

[stapel]

IS MY ANSWER ALSO CORRECT?

As you discovered when you plugged "3x + 1" into the tutor's formulation for "f(x)", no, your answer is not equivalent.

Please reply showing your work and reasoning. Thank you!

Eliz.

 

[Jodene222]

oritinal problem is:

f(x) = 3x + 1, and g(x) = (x^2 + 5x) ^-1/2, fing g (f(x))

when i substitue 3x + 1 for x i get the following:

1
____________________
(3x+1)^2 + 5(3x + 1) this bottom is all under the square root sign, sorry don's know how to type that

then I thought I could multiply it out:

1
___________________________
(9x^2 + 6x + 1) + 15x + 5 all under the square root sign

Now I want to combine like terms:

1
_____________________
9x^2 + 21X + 6 (all in the square root sign)

next I reduced it by taking out 3 to get the following:

1
_____________
3(3x^2 + 7x +2) all under the square root sign.

please advise.
thank you in advance

 

[stapel]

this bottom is all under the square root sign, sorry don's know how to type that

To type mathematical expressions, just use the formatting explained in the links within the "Read Before Posting" thread that you read before posting. :wink:

Given f(x) = 3x + 1, and g(x) = (x^2 + 5x) ^-1/2, find g (f(x))

When I substitue 3x + 1 for x, I get the following:

. . .1/sqrt[(3x+1)^2 + 5(3x + 1)]

. . .1/sqrt[(9x^2 + 6x + 1) + 15x + 5]

. . .1/sqrt[9x^2 + 21x + 6]

Next, I factored inside the radical by taking 3 out front to get the following:

. . .1/sqrt[3(3x^2 + 7x +2)]

Ah. So you meant the "3" to be inside the radical, as well. While this factorization may not be necessary (this will depend upon the text and/or the instructor), your simplification, with the "3" inside the radical, is correct. But if you're going to factor, you should probably factor the quadratic as well.

Eliz.