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Mathematics 14 Online
OpenStudy (anonymous):

help:) f(x) = 3x-1 g(x)= 5/2x+7 h(x)= x^2+1 find: (f•g•h)(x)

OpenStudy (anonymous):

is that the composition symbol?

OpenStudy (anonymous):

(f•g•h)(x) = f( g( h(x)))

OpenStudy (anonymous):

try this notation f(g(h(x)))

OpenStudy (anonymous):

first find fg then subs h in place of x u will get fg(h(x))

OpenStudy (anonymous):

I work from inside to outside...so i start with g(h(x))...this means take the funciton h(x) and use it as the input into g(x).

OpenStudy (anonymous):

\[g(h(x)) = \frac{5}{2(x^{2}+1)+7}=\frac{5}{2x^{2}+9}\]

OpenStudy (anonymous):

yah, i get that part:)

OpenStudy (anonymous):

Now, use this as the input into f(x)

OpenStudy (anonymous):

note i'm assuming its 5/2 * x

OpenStudy (anonymous):

:) SO where are you lost?

OpenStudy (anonymous):

with my final ans.

OpenStudy (anonymous):

\[3( 5/2 * (x^2 +1 ) +7 ) +1\]

OpenStudy (anonymous):

I get \[\frac{15}{2x^{2}+9}-1\]as my final anwer. :)

OpenStudy (anonymous):

based on the previous problems, we used \[g(x)=\frac{5}{2x+7}\]

OpenStudy (anonymous):

i get that too:)

OpenStudy (anonymous):

am i not going to perform the indicated operation 2 subtract it from 1

OpenStudy (anonymous):

See, you don't need our help :)

OpenStudy (anonymous):

I wouldn't. The answer is correct and simplified.

OpenStudy (anonymous):

Anything more is complicating it more than it needs.

OpenStudy (anonymous):

If you were to do that, you'd have to make the 1 into the quadratic on the bottom and then do the subtraction in the numerator.

OpenStudy (anonymous):

That would give you x terms in both top and bottom whereas now you only have them on the bottom...much prettier :)

OpenStudy (anonymous):

so, its better not to subtract it anymore?

OpenStudy (anonymous):

much better not to.

OpenStudy (anonymous):

A commented Mathematica solution is attached.

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