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For the functions f(x)=1/(x)^2  g(x)=7x-6  h(x)=1-2x a) fog(x) c)hof(x) d)gofoh(x)...

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maheen100 | Student, Undergraduate | (Level 1) Honors

Posted June 13, 2012 at 3:35 PM via web

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For the functions f(x)=1/(x)^2  g(x)=7x-6  h(x)=1-2x

a) fog(x)

c)hof(x)

d)gofoh(x)

State restrictions where possible i.e for rational function

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 13, 2012 at 4:19 PM (Answer #1)

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a) You need to compose the functions f(x) and g(x) such that:

`(fog)(x) = f(g(x)) = 1/(g^2(x))`

You need to substitute 7x - 6 for g(x) such that:

`(fog)(x) = f(g(x)) = 1/((7x - 6)^2)`

The denominator of the fraction needs to be different from zero, hence `x!=6/7` .

Thus, the domain of the function needs to exclude the value 6/7.

Hence, evaluating `(fog)(x)`  yields `(fog)(x) = 1/((7x - 6)^2).`

c) You need to compose the functions h(x) and f(x) such that:

`(hof)(x) = h(f(x)) = 1 - 2f(x)`

You need to substitute `1/(x^2)`  for f(x) such that:

`(hof)(x) = h(f(x)) = 1 - 2/(x^2)`

Notice that the domain of the function needs to exclude the value 0.

Hence, evaluating `(hof)(x)`  yields `(hof)(x) = 1 - 2/(x^2).`

d) `(gofoh)(x) = g(f(h(x))) = g(f(1-2x)) = 7f(1-2x) - 6`

`(gofoh)(x) = 7/((1-2x)^2) - 6`

Notice that the domain of the function needs to exclude the value `1/2.`

Hence, evaluating the `(gofoh)(x` ) yields`(gofoh)(x) = 7/((1-2x)^2) - 6`

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