Question

In Exercises \(63-66,\) (a) graph \(f \circ g\) and \(g \circ f\) and make a conjecture about the domain and range of each function. (b) Then confirm your conjectures by finding formulas for \(f \circ g\) and \(g \circ f\) . $$f(x)=x-7, \quad g(x)=\sqrt{x}$$

Short answer

The formulas for \(f \circ g\) and \(g \circ f\) are \(\sqrt{x}-7\) and \(\sqrt{x-7}\) respectively. The domain of \(f \circ g\) is \(x\geq0\) and its range is \(y>-7\). The domain of \(g \circ f\) is \(x\geq7\) and its range is \(y\geq0\).

Step-by-step solution

Find \(f \circ g\)

The composition \(f \circ g\) means \(f(g(x))\). Substituting \(g(x)\) into \(f(x)\) gives \(f(g(x))=f(\sqrt{x})=\sqrt{x}-7\). This implies that wherever you have \(x\) in the function \(f(x)\), it is replaced with \(g(x)=\sqrt{x}\).

Find \(g \circ f\)

The composition \(g \circ f\) means \(g(f(x))\). Substituting \(f(x)\) into \(g(x)\) gives \(g(f(x))=g(x-7)=\sqrt{x-7}\. This implies that wherever you have \(x\) in the function \(g(x)\), it is replaced with \(f(x)=x-7\).

Determine the domains and ranges

For \(f(g(x))=\sqrt{x}-7\), the domain is \(x\geq 0\) because inside the square root, \(x\) must be non-negative. The range is all real numbers greater than \(-7\) as you're minusing \(7\) from square root of \(x\). For \(g(f(x))=\sqrt{x-7}\), the domain is \(x\geq7\) because inside the square root, \(x-7\) must be non-negative. The range is all real numbers greater or equal to \(0\) since the square root function is always non-negative.

Verification by graphing

If you graph these two functions using a graphing calculator, you will see that this analysis of the domain and range matches how the functions look graphically. \(f(g(x))\) will start at \(-7\) and increase for \(x\geq0\) and \(g(f(x))\) will start at \(0\) and increase for \(x\geq7\).