Guided Example #2 - Composition of Functions

Let $(x) = \sqrt{1 - 2 x}$ , and $h (x) = x^{2} + 1$ . Find $(h \circ f) (- 4)$

a. Does not exist
b. 10
c. -4
d. $\sqrt{7}$

The correct answer is B. To see why, view each of the steps below.

By definition, $(h \circ f) (- 4) = h (f (- 4))$ . But $f (- 4) = \sqrt{1 - 2 (- 4)}) = \sqrt{1 + 8} = \sqrt{9} = 3$ .

Therefore, $h (f (- 4)) = h (3) = 3^{2} + 1 = 10$ .

By definition, $(h \circ f) (- 4) = h (f (- 4))$ . But $f (- 4) = \sqrt{1 - 2 (- 4)}) = \sqrt{1 + 8} = \sqrt{9} = 3$ .

Therefore, $h (f (- 4)) = h (3) = 3^{2} + 1 = 10$ .