Checkpoint: Evaluating a Composition of Functions Solution

Use the functions in the previous example to evaluate the composition.

$k [j (- 3)] = k [{(- 3 - 2)}^{3} - 4] k [j (- 3)] = k (- 129) = k (- 129) = - 3 (- 129) + 2 = 389$

$\begin{array}{rcl} [j \circ k] (- a) & = & j [k (- a)] \\ = & j [- 3 (- a) + 2] = j [3 a + 2] \\ = & {((3 a + 2) - 2)}^{3} - 4 = {(3 a)}^{3} - 4 \\ [j \circ k] (- a) & = & 27 a^{3} - 4 \end{array}$

Note

Using a calculator for these problems is more efficient than calculating by hand.

Q: Why is the solution to “the function j composed with k of 2” $j [k (2)]$ not –4?

A: Because the intersection point is not the composition of the functions.