Math Functions Practice with Answer Key

1. B. 10x + 27

f(x) = 2x + 5
g(x) = 5x + 2.
g○f = g(f(x)) = g(2x + 5) = 5(2x + 5) + 2 = 10x + 25 + 2 = 10x + 27

2. D. 2+2×2+x4
f(x)=1+x2
f○f=f(f(x))=f(1+x2)=1+(1+x2)2=1+1+2×2+x4=2+2×2+x4

3. B. -2x² – 1
f(x) = -x
g(x) = 2x + 1
h(x) = x²
f○g○h = f(g(h(x))) = f(g(x²)) = f(2x² + 1) = -(2x² + 1) = -2x² – 1

3. C

Without any restrictions, the domain of a function is (-∞, +∞). The restrictions are found by checking the denominator of the function. If there are any values that make the denominator zero; since division by zero is undefined, these x values should be eliminated from the domain:

f(x) = (x + 1) / (x – 2) : Notice that the denominator is x – 2 and x = 2 value makes it zero, so makes the function undefined. The domain of this function is (-∞, +∞) – {2}.

f(x) = (x + 7) / (x² + 5x + 6) : Notice that the denominator is x² + 5x + 6 which can be factorised as

(x + 2)(x + 3). Then, x = – 2 and x = – 3 values make the denominator zero, so make the function undefined. The domain of this function is (-∞, +∞) – {- 3, – 2}.

f(x) = (x² – 9) / (x + 3) : Notice that first, it is possible to simplify the function:

f(x) = (x² – 9) / (x + 3) = (x + 3)(x – 3) / (x + 3) = x – 3. Then, the denominator has vanished; the domain of this function is (-∞, +∞).

f(x) = (4x + 7) / (9x² – 4) : Notice that the denominator is 9x² – 4 which can be factorised as

(3x – 2)(3x + 2). It is not possible to simplify. Then, x = 2/3 and x = – 2/3 values make the denominator zero, so make the function undefined. The domain of this function is (-∞, +∞) – {- 2/3, 2/3}.

Function (x² – 9) / (x + 3) has the largest domain.

The correct answer is (C).