12 multiple-choice questions, progressively harder.
Let f(x)=x2f(x) = x^2f(x)=x2 and g(x)=xg(x) = \sqrt{x}g(x)=x. What is (g∘f)(x)(g \circ f)(x)(g∘f)(x)?
Solution
Correct answer: C
Substitute f(x)=x2f(x) = x^2f(x)=x2 into g(x)=xg(x) = \sqrt{x}g(x)=x.
(g∘f)(x)=x2=∣x∣(g \circ f)(x) = \sqrt{x^2} = \lvert x \rvert(g∘f)(x)=x2=∣x∣
The square root of a square is the absolute value, not xxx; for x=−3x = -3x=−3 it gives 333, not −3-3−3.
Let f(x)=x2f(x) = x^2f(x)=x2 and g(x)=xg(x) = \sqrt{x}g(x)=x. What is the domain of g∘fg \circ fg∘f?
Correct answer: D
The inner value x2x^2x2 is always nonnegative, so x2\sqrt{x^2}x2 is always defined.
x2≥0 for every real xx^2 \ge 0 \text{ for every real } xx2≥0 for every real x
So ggg accepts every output of fff, and the domain is all real numbers.
Let f(x)=xf(x) = \sqrt{x}f(x)=x and g(x)=x2g(x) = x^2g(x)=x2. Which statement about g∘fg \circ fg∘f is TRUE?
Compute the rule and its domain.
(g∘f)(x)=(x)2=x,x≥0(g \circ f)(x) = (\sqrt{x})^2 = x, \qquad x \ge 0(g∘f)(x)=(x)2=x,x≥0
The formula is xxx, but the domain x≥0x \ge 0x≥0 is smaller than the identity's, so by the equality test they are different functions.
Which of these is NOT a valid decomposition of h(x)=(5x−2)3h(x) = (5x - 2)^3h(x)=(5x−2)3 as g∘fg \circ fg∘f?
Compose each and compare with (5x−2)3(5x - 2)^3(5x−2)3.
g(f(x))=3(5x−2)=15x−6≠(5x−2)3for f(x)=5x−2, g(x)=3xg(f(x)) = 3(5x - 2) = 15x - 6 \ne (5x - 2)^3 \quad\text{for } f(x) = 5x - 2,\ g(x) = 3xg(f(x))=3(5x−2)=15x−6=(5x−2)3for f(x)=5x−2, g(x)=3x
The other three each compose to (5x−2)3(5x - 2)^3(5x−2)3, so they are valid decompositions.
Which choice is a valid decomposition of h(x)=x2+4h(x) = \sqrt{x^2 + 4}h(x)=x2+4 as g∘fg \circ fg∘f?
Correct answer: A
A valid split needs g(f(x))=x2+4g(f(x)) = \sqrt{x^2 + 4}g(f(x))=x2+4.
g(f(x))=x2+4whenf(x)=x2+4, g(x)=xg(f(x)) = \sqrt{x^2 + 4} \quad\text{when}\quad f(x) = x^2 + 4,\ g(x) = \sqrt{x}g(f(x))=x2+4whenf(x)=x2+4, g(x)=x
The others give x+4x + 4x+4, ∣x∣+4\lvert x \rvert + 4∣x∣+4, and ∣x+4∣\lvert x + 4 \rvert∣x+4∣.
Let f(x)=x2−2f(x) = x^2 - 2f(x)=x2−2. Using f2=f∘ff^2 = f \circ ff2=f∘f, what is f2(x)f^2(x)f2(x)?
f2=f∘ff^2 = f \circ ff2=f∘f, so substitute f(x)=x2−2f(x) = x^2 - 2f(x)=x2−2 into fff.
f2(x)=(x2−2)2−2=x4−4x2+4−2=x4−4x2+2f^2(x) = (x^2 - 2)^2 - 2 = x^4 - 4x^2 + 4 - 2 = x^4 - 4x^2 + 2f2(x)=(x2−2)2−2=x4−4x2+4−2=x4−4x2+2
The output square (f(x))2=x4−4x2+4(f(x))^2 = x^4 - 4x^2 + 4(f(x))2=x4−4x2+4 is a different function.
Let f(x)=x+3f(x) = x + 3f(x)=x+3 and g(x)=x2g(x) = x^2g(x)=x2. What is (f∘g)(x)−(g∘f)(x)(f \circ g)(x) - (g \circ f)(x)(f∘g)(x)−(g∘f)(x)?
Correct answer: B
Compute both orders and subtract.
(f∘g)(x)−(g∘f)(x)=(x2+3)−(x+3)2=−6x−6(f \circ g)(x) - (g \circ f)(x) = (x^2 + 3) - (x + 3)^2 = -6x - 6(f∘g)(x)−(g∘f)(x)=(x2+3)−(x+3)2=−6x−6
The difference −6x−6-6x - 6−6x−6 is zero only at x=−1x = -1x=−1, so the compositions differ.
Write y=−f(x)+4y = -f(x) + 4y=−f(x)+4 as A∘fA \circ fA∘f. What is the output map A(y)A(y)A(y)?
Negating then adding 444 acts on the output y=f(x)y = f(x)y=f(x).
y=−f(x)+4=(A∘f)(x),A(y)=−y+4y = -f(x) + 4 = (A \circ f)(x), \quad A(y) = -y + 4y=−f(x)+4=(A∘f)(x),A(y)=−y+4
The reflection negates the height and the +4+4+4 raises it.
Let f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 and g(x)=x2−3g(x) = x^2 - 3g(x)=x2−3. What is (g∘f)(2)(g \circ f)(2)(g∘f)(2)?
Apply fff first, then ggg.
f(2)=5,g(5)=52−3=22f(2) = 5, \qquad g(5) = 5^2 - 3 = 22f(2)=5,g(5)=52−3=22
So (g∘f)(2)=22(g \circ f)(2) = 22(g∘f)(2)=22; forgetting the −3-3−3 gives the trap 252525.
Let f(x)=xf(x) = \sqrt{x}f(x)=x. Using f2=f∘ff^2 = f \circ ff2=f∘f, what is the domain of f2f^2f2?
f2(x)=xf^2(x) = \sqrt{\sqrt{x}}f2(x)=x needs the inner root defined and its output accepted by the outer root.
x≥0 and x≥0 (always) ⟺ x≥0x \ge 0 \text{ and } \sqrt{x} \ge 0 \text{ (always)} \iff x \ge 0x≥0 and x≥0 (always)⟺x≥0
The outer root adds no new restriction since x≥0\sqrt{x} \ge 0x≥0 already, so the domain is x≥0x \ge 0x≥0.
Let f(x)=x+2f(x) = x + 2f(x)=x+2. What is (f∘f∘f)(x)(f \circ f \circ f)(x)(f∘f∘f)(x)?
Apply fff three times, each adding 222.
f3(x)=((x+2)+2)+2=x+6f^3(x) = ((x + 2) + 2) + 2 = x + 6f3(x)=((x+2)+2)+2=x+6
Three shifts by 222 add to a shift by 666.
How do (h∘g)∘f(h \circ g) \circ f(h∘g)∘f and h∘(g∘f)h \circ (g \circ f)h∘(g∘f) compare?
Both groupings send xxx to the same value.
((h∘g)∘f)(x)=h(g(f(x)))=(h∘(g∘f))(x)\big((h \circ g) \circ f\big)(x) = h(g(f(x))) = \big(h \circ (g \circ f)\big)(x)((h∘g)∘f)(x)=h(g(f(x)))=(h∘(g∘f))(x)
They share domain and values, so they are always equal; that is associativity, and it needs no commuting.
Reset this practice set?
This clears every answer you have given and starts the set again from question 1.