Question

1 х Given g(x)=


`\sqrt[3]{x}`
and h(x) =

`\frac{1}{ {x}^(3) }`
(a) Find f(x) such that (fogoh)(x)=

`(x)/(x + 1)`
Determine the domain of (fogoh) (x)​

Answer 1

(a) Since
`g(x)=\sqrt[3]{x}` and
`h(x) = \frac1{x^3}`, we have

`(g\circ h)(x) = g(h(x)) = g\left(\frac1{x^3}\right) = √(3){\frac1{x^3}} = \frac1x`

We're given that

`(f \circ g \circ h)(x) = f(g(h(x))) = f\left(\frac1x\right) = \frac x{x+1}`

but we can rewrite this as

`\frac x{x+1} = (\frac xx)/(\frac xx + \frac1x) = \frac1{1+\frac1x}`

(bear in mind that we can only do this so long as x ≠ 0) so it follows that

`f\left(\frac1x\right) = \frac1{1+\frac1x} \implies \boxed{f(x) = \frac1{1+x}}`

(b) On its own, we may be tempted to conclude that the domain of
`(f\circ g\circ h)(x) = \frac1{1+x}` is simply x ≠ -1. But we should be more careful. The domain of a composite depends on each of the component functions involved.

`g(x) = \sqrt[3]{x}` is defined for all x - no issue here.

`h(x) = \frac1{x^3}` is defined for all x ≠ 0. Then
`(g\circ h)(x) = \frac1x` also has a domain of x ≠ 0.

`f(x) = \frac1{1+x}` is defined for all x ≠ -1, but

`(f\circ g\circ h)(x)=f\left(\frac1x\right) = \frac1{1+\frac1x}`

is undefined not only at x = -1, but also at x = 0. So the domain of
`(f\circ g\circ h)(x)` is

`\left\{x\in\mathbb R \mid x\\eq-1 \text{ and }x\\eq0\right\}`