🎓 BookMCQ
← Back to 1. Basics before calculus

📝 Expressing functions as compositions (14 MCQs)

📖 From Calculus • 1. Basics before calculus • 14 questions available

What is Expressing functions as compositions?

Definition:
Expressing a function as a composition means decomposing a given function into two or more simpler functions, where the original function equals f(g(x))f(g(x)) or f(g(h(x)))f(g(h(x))), to simplify analysis or computation.

Example:
For h(x)=sin(3x2)h(x) = \sin(3x^2), let g(x)=3x2g(x) = 3x^2 and f(u)=sin(u)f(u) = \sin(u), then h(x)=f(g(x))h(x) = f(g(x)).

Reason:
This decomposition aids in differentiation using the chain rule, integration by substitution, and understanding step-by-step transformations in complex functions.

4
Easy
6
Medium
4
Hard

📝 All Expressing functions as compositions MCQs

Q1. Which function serves as the inside function \g(x)\ when the function \h(x)=\\sqrt{x+4}\ is written as a composition \h(x)=f(g(x))\?

A.\g(x)=\\sqrt{x}\
B.\g(x)=x+4\
C.\g(x)=\\frac{x}{4}\
D.\g(x)=x^{2}+4\
💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The expression \h(x)=\\sqrt{x+4}\ can be written as \f(g(x))\ with \g(x)\ providing the quantity to be square‑rooted. Here \g(x)=x+4\ supplies the argument of the outer square‑root function \f(u)=\\sqrt{u}\, making option B correct.

Q2. If \h(x)=(3x-2)^{5}\ is expressed as \h(x)=f(g(x))\, which of the following statements about the monotonicity of \f\ and \g\ is necessarily true?

A.Both \f\ and \g\ are increasing on \\\mathbb{R}\
B.\f\ is increasing while \g\ is decreasing
C.\f\ is decreasing while \g\ is increasing
D.Neither \f\ nor \g\ is monotonic on \\\mathbb{R}\
💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The inner function \g(x)=3x-2\ is a linear function with positive slope, so it is strictly increasing. The outer function \f(u)=u^{5}\ is also strictly increasing because odd powers preserve order. Hence both are increasing, giving option A.

Q3. Which decomposition of \(x^{2}+1)^{10}\ yields an inside function that is a polynomial of lower degree than the other options?

A.\g(x)=(x^{2}+1)^{2},\\; f(x)=x^{5}\
B.\g(x)=(x^{2}+1)^{5},\\; f(x)=x^{2}\
C.\g(x)=x^{2}+1,\\; f(x)=x^{10}\
D.\g(x)=(x^{2}+1)^{3},\\; f(x)=x^{10/3}\
💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Choosing \g(x)=x^{2}+1\ gives an inside function of degree 2, which is lower than the degrees in the other choices where the inside function contains higher powers of \(x^{2}+1)\. Thus option C is correct.

Q4. Express \\\sin(x^{3}+2)\ as a composition \f(g(x))\. Which pair \(f,g)\ is correct?

A.\f(x)=\\sin x,\\; g(x)=x^{3}+2\
B.\f(x)=\\sin(x+2),\\; g(x)=x^{3}\
C.\f(x)=\\sin x,\\; g(x)=x+2\
D.\f(x)=\\sin x,\\; g(x)=x^{3}\
💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: To obtain \\\sin(x^{3}+2)\ we first compute \x^{3}+2\ (the inside step) and then apply the sine function. Hence \g(x)=x^{3}+2\ and \f(u)=\\sin u\, which corresponds to option A.

Q5. For \h(x)=\\ln(2x+1)\ written as \h(x)=f(g(x))\, which statement about the domain of the inside function \g\ is necessarily true?

A.\g\ is defined for all real numbers.
B.\g(x)=2x+1\ must be positive for all \x\.
C.The domain of \g\ is \\\{x\\in\\mathbb{R}\\mid 2x+1>0\\}\. ✅
D.\g\ has no restrictions because the logarithm handles negatives.
💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The logarithm requires its argument to be positive, so the inside function \g(x)=2x+1\ must satisfy \2x+1>0\. This translates to the domain \x>-\\tfrac{1}{2}\, which is exactly option C.

Q6. Consider \h(x)=e^{(x-1)^{2}}\. Which of the following decompositions makes the outer function linear in its argument?

A.\g(x)=x-1,\\; f(x)=e^{x^{2}}\
B.\g(x)=(x-1)^{2},\\; f(x)=e^{x}\
C.\g(x)=e^{x-1},\\; f(x)=x^{2}\
D.\g(x)=e^{(x-1)^{2}},\\; f(x)=x\
💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: If we let \g(x)=e^{(x-1)^{2}}\ and choose \f(u)=u\ (the identity function), the outer function is linear (of the form \f(u)=u\). This satisfies the requirement, making option D correct.

Q7. Which pair \(f,g)\ correctly represents \h(x)=\\frac{1}{x^{2}+3}\ as a composition \h(x)=f(g(x))\?

A.\g(x)=\\frac{1}{x^{2}+3},\\; f(x)=x\
B.\g(x)=x^{2}+3,\\; f(x)=\\frac{1}{x}\
C.\g(x)=x,\\; f(x)=\\frac{1}{x^{2}+3}\
D.\g(x)=x^{2},\\; f(x)=\\frac{1}{x+3}\
💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Writing \h(x)=\\frac{1}{x^{2}+3}\ as \f(g(x))\ suggests letting the inner function produce the denominator, i.e., \g(x)=x^{2}+3\, and the outer function take the reciprocal, \f(u)=\\frac{1}{u}\. This matches option B.

Q8. If \h = f\\circ g\ and both \f\ and \g\ are one‑to‑one functions, which statement must hold for \h\?

A.\h\ is also one‑to‑one. ✅
B.\h\ is onto.
C.\h\ is constant.
D.No conclusion can be drawn about \h\.
💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The composition of two injective (one‑to‑one) functions is itself injective. Since each input maps to a unique output through \g\ and then uniquely through \f\, the overall map \h\ is one‑to‑one, giving option A.

Q9. For \h(x)=\\tan(\\sqrt{x})\, which decomposition correctly identifies the inside function as a root function?

A.\g(x)=\\tan x,\\; f(x)=\\sqrt{x}\
B.\g(x)=\\sqrt{x},\\; f(x)=\\tan x\
C.\g(x)=x^{1/2},\\; f(x)=\\tan x\
D.\g(x)=\\tan(\\sqrt{x}),\\; f(x)=x\
💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The expression \\\tan(\\sqrt{x})\ is evaluated by first taking the square root of \x\ (a root function) and then applying the tangent. Hence \g(x)=\\sqrt{x}\ and \f(u)=\\tan u\, which is option B.

Q10. When differentiating \h(x)=(5x+7)^{4}\ using the chain rule, which inside function simplifies the derivative most effectively?

A.\g(x)=5x+7\
B.\g(x)=(5x+7)^{2}\
C.\g(x)=x\
D.\g(x)=5x+7\ (duplicate)
💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The chain rule works best when the inner function is the linear expression that appears inside the power, i.e., \g(x)=5x+7\. The outer function then becomes \f(u)=u^{4}\, leading to a straightforward derivative; option A captures this.

Q11. Under what condition does swapping the order of composition, i.e., defining \k(x)=g(f(x))\, yield the same function as \h(x)=f(g(x))\?

A.When \f\ and \g\ are inverses of each other.
B.When both \f\ and \g\ are constant functions.
C.When \f\ and \g\ are linear with the same slope.
D.When \f\ and \g\ commute, i.e., \f(g(x))=g(f(x))\ for all \x\. ✅
💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Two functions commute if applying them in either order gives the same result for every input. This is precisely the condition needed for \k(x)=g(f(x))\ to equal \h(x)=f(g(x))\; thus option D is correct.

Q12. Two students decompose \h(x)=\\sqrt{(2x+1)^{2}+4}\. Which decomposition results in the outer function being a square root of a sum?

A.\g(x)=(2x+1)^{2},\\; f(u)=\\sqrt{u+4}\
B.\g(x)=2x+1,\\; f(u)=\\sqrt{u^{2}+4}\
C.\g(x)=(2x+1)^{2}+4,\\; f(u)=\\sqrt{u}\
D.\g(x)=\\sqrt{(2x+1)^{2}+4},\\; f(u)=u\
💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: If we let the inside function produce the entire radicand \(2x+1)^{2}+4\, then the outer function is simply the square root of that quantity. This matches option C, where \g(x)=(2x+1)^{2}+4\ and \f(u)=\\sqrt{u}\.

Q13. Which pair \(f,g)\ expresses \h(x)=\\cos(2x-\\pi)\ as a composition \f(g(x))\?

A.\f(u)=\\cos(2u),\\; g(x)=x-\\pi\
B.\f(u)=\\cos u,\\; g(x)=2x-\\pi\
C.\f(u)=2\\cos u,\\; g(x)=x-\\pi\
D.\f(u)=\\cos(u-\\pi),\\; g(x)=2x\
💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The function inside the cosine is \2x-\\pi\. Taking this as the inside function \g(x)=2x-\\pi\ and the outer function as the cosine itself, \f(u)=\\cos u\, reproduces \h(x)\. Hence option B is correct.

Q14. For \h(x)=\\frac{(x^{2}-1)^{3}}{(x-1)^{2}}\, which composition \f(g(x))\ eliminates the removable singularity at \x=1\ by simplifying the expression before applying the outer function?

A.\g(x)=\\frac{x+1}{x-1},\\; f(u)=u^{3}\
B.\g(x)=x^{2}-1,\\; f(u)=\\frac{u^{3}}{(x-1)^{2}}\
C.\g(x)=\\frac{x^{2}-1}{x-1},\\; f(u)=\\frac{u^{3}}{x-1}\
D.\g(x)=x+1,\\; f(u)=u^{3}(x-1)\
💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Factor \(x^{2}-1)=(x-1)(x+1)\. Cancelling one \(x-1)\ factor gives \h(x)=(x-1)(x+1)^{3}\. Writing \g(x)=\\frac{x+1}{x-1}\ captures the ratio that removes the singularity, and then \f(u)=u^{3}\ restores the cubic power, matching option A.

🔗 Related Topics (MCQs)