Problem 6 - Calculus of a Single Variable

Question

Show that \(f\) and \(g\) are inverse functions (a) analytically and (b) graphically. \(f(x)=16-x^{2}, \quad x \geq 0, \quad g(x)=\sqrt{16-x}\)

Step-by-Step Solution

Answer

Analytically, both the composite functions \(f(g(x))\) and \(g(f(x))\) equal to \(x\), showing that \(f\) and \(g\) are inverses of each other. Graphically, the graph of \(g(x)\) is a reflection of the graph of \(f(x)\) over the line \(y=x\), which also shows that \(f\) and \(g\) are inverses of each other.

Step 1: Analytical Proof: Step 1

First, let's find \(f(g(x))\). We plug \(g(x)\) into \(f(x)\), so we get: \(f(g(x)) = f(\sqrt{16-x})\). Expanding, this gives us: \(f(\sqrt{16-x}) = 16 - (\sqrt{16-x})^{2}\).

Step 2: Analytical Proof: Step 2

Since \((\sqrt{16-x})^{2}\) is simply \(16-x\), the expression simplifies to: \(f(g(x)) = 16 - (16 - x) = x\). Thus, \(f(g(x))=x\), which is one half of the condition for \(f\) and \(g\) to be inverse functions.

Step 3: Analytical Proof: Step 3

Now let's find \(g(f(x))\). We plug \(f(x)\) into \(g(x)\), to get: \(g(f(x)) = g(16-x^{2})\). Expanding, this gives us: \(g(16-x^{2}) = \sqrt{16 - (16 - x^{2})}\).

Step 4: Analytical Proof: Step 4

The expression under the square root simplifies to \(x^{2}\), so our expression now is: \(g(f(x)) = \sqrt{x^2}\). Since \(x \geq 0\), \(\sqrt{x^{2}}\) equals \(x\) and our current expression simplifies to: \(g(f(x)) = x\). Therefore, both \(f(g(x))=x\) and \(g(f(x))=x\), satisfying the conditions for \(f\) and \(g\) to be inverses of each other analytically.

Step 5: Graphical Proof: Step 1

To prove the functions are inverses graphically, start by sketching the graph of \(y=f(x)\). This is a downward-opening parabola centered at \(x=0\) and with a vertex at (0,16). The restriction on \(x \geq 0\) makes it half of a full parabola.

Step 6: Graphical Proof: Step 2

Sketch the graph of \(y=g(x)\). This is a sideways-opening parabola (actually half of one since the range is restricted to \(y \geq 0\)). It opens to the right and is centered on \(y=0\) with a vertex at (16,0).

Step 7: Graphical Proof: Step 3

Sketch the line \(y=x\), which is the line of reflection for inverse functions. If \(f(x)\) and \(g(x)\) are indeed inverses, the reflection of \(f(x)\) over \(y=x\) should match the graph of \(g(x)\). Once you see that this is the case, you will have completed the proof graphically.