Find the composite function, domain, range, and solve a composition equation

Expert intro: Composite functions can look messy at first, but most of the work is just keeping track of what goes inside what, and then checking the allowed inputs carefully.

Detailed walkthrough

We are given

$f(x)=\sqrt{x+4},\qquad x\ge -3$

and

$g(x)=2x^2-3,\qquad x\le 47.$

a) Find a simplified expression for $g(f(x))$

Substitute $f(x)$ into $g$:

$g(f(x))=2\left(\sqrt{x+4}\right)^2-3.$

Since the square root is squared,

$g(f(x))=2(x+4)-3=2x+5.$

b) Domain and range of $g(f(x))$

Domain

For $f(x)=\sqrt{x+4}$, we need

$x+4\ge 0 \Rightarrow x\ge -4.$

The given extra condition $x\ge -3$ is stricter, so the domain of $f$ is

$[-3,\infty).$

Now check whether the output of $f$ fits into the domain of $g$, which requires input $\le 47$. Since

$f(x)=\sqrt{x+4},$

its values are nonnegative, and $\sqrt{x+4}\le 47$ is true for all $x\ge -3$ here. So the composite is defined for

$\boxed{[-3,\infty)}.$

Range

Since

$g(f(x))=2x+5$

and $x\ge -3$, the smallest value occurs at $x=-3$:

$2(-3)+5=-1.$

As $x$ increases, $2x+5$ increases without bound. So the range is

$\boxed{[-1,\infty)}.$

c) Solve $f(g(x))=17$

Compute the composition:

$f(g(x))=\sqrt{g(x)+4}=\sqrt{2x^2-3+4}=\sqrt{2x^2+1}.$

Set equal to 17:

$\sqrt{2x^2+1}=17.$

Square both sides:

$2x^2+1=289$

$2x^2=288$

$x^2=144$

$x=\pm 12.$

Now check the domain restriction for $g$: $x\le 47$, so both values work.

Final answers

$\boxed{g(f(x))=2x+5}$

$\boxed{\text{Domain }=[-3,\infty)}$

$\boxed{\text{Range }=[-1,\infty)}$

$\boxed{x=\pm 12}$

💡 Pitfall guide

A very common mistake is to forget that composition domain depends on both functions. You cannot stop after checking only $f$'s domain; you also need the output of $f$ to fit the input rule for $g$. Another easy slip is squaring $\sqrt{2x^2+1}=17$ and then forgetting the negative square root is not allowed here.

🔄 Real-world variant

If the domain restriction on $f$ had been $x\ge -4$ instead of $x\ge -3$, the domain of $g(f(x))$ would still be $[-4,\infty)$, but the range would change to $[ -3,\infty )$. If the equation were $g(f(x))=17$, then you would solve

$2x+5=17$

which gives $x=6$, and then check that $6$ fits the domain.

🔍 Related terms

function composition, domain of composite, range of a function