Question

1 х Given g(x)=

Answer 1

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

$(g\circ h)(x) = g(h(x)) = g\left(\dfrac1{x^3}\right) = \sqrt{3}{\dfrac1{x^3}} = \dfrac1x$

We're given that

$(f \circ g \circ h)(x) = f(g(h(x))) = f\left(\dfrac1x\right) = \dfrac x{x+1}$

but we can rewrite this as

$\dfrac x{x+1} = \dfrac{\frac xx}{\frac xx + \frac1x} = \dfrac1{1+\frac1x}$

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

$f\left(\dfrac1x\right) = \dfrac1{1+\frac1x} \implies \boxed{f(x) = \dfrac1{1+x}}$

(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

$(f\circ g\circ h)(x)=f\left(\frac1x\right) = \dfrac1{1+\frac1x}$

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

$\left\{x\in\mathbb R \mid x\neq-1 \text{ and }x\neq0\right\}$