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Algebra Help: Combine Different Function Types

Review real example questions for Combine Different Function Types in Algebra.

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Let f(x)=x2+2f(x)=x^2+2 and g(x)=x−1g(x)=x-1. Find (f⋅g)(x)(f\cdot g)(x).

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Question 1

Let f(x)=x2+2f(x)=x^2+2 and g(x)=x−1g(x)=x-1. Find (f⋅g)(x)(f\cdot g)(x).

  1. (f⋅g)(x)=x2+2x−1(f\cdot g)(x)=x^2+2x-1
  2. (f⋅g)(x)=x3+x2+2x−2(f\cdot g)(x)=x^3+x^2+2x-2
  3. (f⋅g)(x)=x3−x2+2x−2(f\cdot g)(x)=x^3-x^2+2x-2 (correct answer)
  4. (f⋅g)(x)=x2+x+1(f\cdot g)(x)=x^2+x+1

Explanation: This question tests your ability to combine standard function types—like linear, quadratic, exponential, and constant functions—using arithmetic operations to build more complex functions. Combining different function types creates new behaviors: adding a constant to an exponential (like y = 2x2^x + 3) shifts the exponential curve vertically, adding a linear to a quadratic changes the parabola's shape, and so on. Each function type brings its characteristic behavior to the combination! To find (f · g)(x) where f(x) = x² + 2 and g(x) = x - 1, we multiply the expressions: (f · g)(x) = (x² + 2)(x - 1) = x²(x - 1) + 2(x - 1) = x³ - x² + 2x - 2. Simplifying: x³ - x² + 2x - 2, a new cubic function combining quadratic and linear. Choice A correctly combines the functions by multiplying and expanding to x³ - x² + 2x - 2. Choice B has a sign error when multiplying: in (x² + 2)(x - 1), the x² * (-1) = -x², not +x²—watch the distribution! The recipe for combining functions: (1) Write out what (f · g)(x) means: f(x) · g(x), (2) Substitute the actual function expressions, (3) Perform the operation carefully, (4) Simplify by combining like terms. Common pitfall: don't confuse (f · g)(x) with f(g(x)). The notation (f · g)(x) means 'multiply the functions,' while f(g(x)) means 'composition' (plug g into f). If you see +, -, ×, or ÷ in the notation, you're combining outputs, not composing functions!

Question 2

Let f(x)=∣x∣f(x)=|x| (absolute value) and g(x)=x−4g(x)=x-4 (linear). Simplify (f+g)(x)(f+g)(x).

  1. (f+g)(x)=∣x∣−(x−4)(f+g)(x)=|x|-(x-4)
  2. (f+g)(x)=∣x−4∣(f+g)(x)=|x-4|
  3. (f+g)(x)=∣x∣(x−4)(f+g)(x)=|x|(x-4)
  4. (f+g)(x)=∣x∣+x−4(f+g)(x)=|x|+x-4 (correct answer)

Explanation: This question tests your ability to combine standard function types—like linear, quadratic, exponential, and constant functions—using arithmetic operations to build more complex functions. When we combine functions using addition, we write (f+g)(x)=f(x)+g(x)(f + g)(x) = f(x) + g(x), which means 'add the outputs': for any input x, evaluate both f(x) and g(x), then add those results. For example, if f(x)=2x+1f(x) = 2x + 1 and g(x)=x2g(x) = x^2, then (f+g)(x)=(2x+1)+x2=x2+2x+1(f + g)(x) = (2x + 1) + x^2 = x^2 + 2x + 1. The result is a new function that combines both behaviors! To find (f+g)(x)(f + g)(x) where f(x)=∣x∣f(x) = |x| and g(x)=x−4g(x) = x - 4, we add the expressions: (f+g)(x)=∣x∣+(x−4)=∣x∣+x−4(f + g)(x) = |x| + (x - 4) = |x| + x - 4. Simplifying: it can't be combined further without piecewise definition, but this form shows the absolute value plus linear. This new function combines the V-shaped absolute value with the linear shift. Choice B correctly combines the functions by adding the absolute value and linear terms, giving (f+g)(x)=∣x∣+x−4(f + g)(x) = |x| + x - 4. Choice C confuses addition with multiplication: (f+g)(x)(f + g)(x) means add the outputs, NOT multiply. These are very different operations! The recipe for combining functions: (1) Write out what (f+g)(x)(f + g)(x) means: f(x)+g(x)f(x) + g(x), (2) Substitute the actual function expressions, (3) Perform the operation carefully, (4) Simplify by combining like terms. Common pitfall: don't confuse (f+g)(x)(f + g)(x) with f(g(x))f(g(x)). The notation (f+g)(x)(f + g)(x) means 'add the functions,' while f(g(x))f(g(x)) means 'composition' (plug g into f). If you see +, -, ×, or ÷ in the notation, you're combining outputs, not composing functions!

Question 3

A ride-share company models the total fare (in dollars) as the sum of a fixed booking fee and a distance-based charge. Let f(x)=2.50f(x)=2.50 (a constant booking fee) and g(x)=1.75xg(x)=1.75x (a linear distance charge for xx miles). Write the combined function h(x)=(f+g)(x)h(x)=(f+g)(x).

  1. h(x)=4.25xh(x)=4.25x
  2. h(x)=1.75x−2.50h(x)=1.75x-2.50
  3. h(x)=2.50+1.75xh(x)=2.50+1.75x (correct answer)
  4. h(x)=2.50(1.75x)h(x)=2.50(1.75x)

Explanation: This question tests your ability to combine standard function types—like linear, quadratic, exponential, and constant functions—using arithmetic operations to build more complex functions. When we combine functions using addition, we write (f + g)(x) = f(x) + g(x), which means 'add the outputs': for any input x, evaluate both f(x) and g(x), then add those results. For example, if f(x) = 2x + 1 and g(x) = x², then (f + g)(x) = (2x + 1) + x² = x² + 2x + 1. The result is a new function that combines both behaviors! To find (f + g)(x) where f(x) = 2.50 and g(x) = 1.75x, we add the expressions: (f + g)(x) = 2.50 + 1.75x, which is already in its simplest form as a linear function with a constant term. This new function combines the constant booking fee with the linear distance charge to model the total fare. Choice C correctly combines the functions by adding the constant and linear terms, giving h(x) = 2.50 + 1.75x. Choice D attempts to multiply but uses the wrong operation: the question asks for addition, not multiplication. Make sure you're performing the operation that's actually requested! The recipe for combining functions: (1) Write out what (f + g)(x) means: f(x) + g(x), (2) Substitute the actual function expressions, (3) Perform the operation carefully, (4) Simplify by combining like terms. Real-world combinations make intuitive sense: total fare = booking fee + distance charge (add), which matches how we naturally combine quantities!

Question 4

A projectile's height (in meters) is modeled as a quadratic term plus a constant initial height. Let f(t)=−5t2+20tf(t)=-5t^2+20t and g(t)=2g(t)=2. Write the combined function h(t)=(f+g)(t)h(t)=(f+g)(t).​

  1. h(t)=−5(t+2)2+20th(t)=-5(t+2)^2+20t
  2. h(t)=−5t2+22th(t)=-5t^2+22t
  3. h(t)=(−5t2+20t)(2)h(t)=(-5t^2+20t)(2)
  4. h(t)=−5t2+20t+2h(t)=-5t^2+20t+2 (correct answer)

Explanation: This question tests your ability to combine standard function types—like linear, quadratic, exponential, and constant functions—using arithmetic operations to build more complex functions. When we combine functions using addition, we write (f + g)(x) = f(x) + g(x), which means 'add the outputs': for any input x, evaluate both f(x) and g(x), then add those results. For example, if f(x) = 2x + 1 and g(x) = x², then (f + g)(x) = (2x + 1) + x² = x² + 2x + 1. The result is a new function that combines both behaviors! To find (f + g)(t) where f(t) = -5t² + 20t and g(t) = 2, we add the expressions: (f + g)(t) = -5t² + 20t + 2, which is already in quadratic form. This new function combines the quadratic projectile motion with the constant initial height. Choice A correctly combines the functions by adding the quadratic and constant, giving h(t) = -5t² + 20t + 2. Choice C confuses addition with multiplication: (f + g)(t) means add the outputs, NOT multiply. These are very different operations! The recipe for combining functions: (1) Write out what (f + g)(x) means: f(x) + g(x), (2) Substitute the actual function expressions, (3) Perform the operation carefully, (4) Simplify by combining like terms. Combining different function types creates new behaviors: adding a constant to an exponential (like y = 2x2^x + 3) shifts the exponential curve vertically, adding a linear to a quadratic changes the parabola's shape, and so on. Each function type brings its characteristic behavior to the combination!

Question 5

Let f(x)=x2−4x+1f(x)=x^2-4x+1 and g(x)=3x−5g(x)=3x-5. Find (f+g)(x)(f+g)(x).

  1. (f+g)(x)=x2−4x+1+(3x−5)(f+g)(x)=x^2-4x+1+(3x-5)
  2. (f+g)(x)=x2+12x−5(f+g)(x)=x^2+12x-5
  3. (f+g)(x)=x2−7x+6(f+g)(x)=x^2-7x+6
  4. (f+g)(x)=x2−x−4(f+g)(x)=x^2-x-4 (correct answer)

Explanation: This question tests your ability to combine standard function types—like linear, quadratic, exponential, and constant functions—using arithmetic operations to build more complex functions. When we combine functions using addition, we write (f + g)(x) = f(x) + g(x), which means 'add the outputs': for any input x, evaluate both f(x) and g(x), then add those results. For example, if f(x) = 2x + 1 and g(x) = x², then (f + g)(x) = (2x + 1) + x² = x² + 2x + 1. The result is a new function that combines both behaviors! To find (f + g)(x) where f(x) = x² - 4x + 1 and g(x) = 3x - 5, we add the expressions: (f + g)(x) = (x² - 4x + 1) + (3x - 5) = x² - 4x + 3x + 1 - 5. Simplifying: x² - x - 4, a new function that combines quadratic and linear behaviors into another quadratic. Choice B correctly combines the functions by adding and simplifying to x² - x - 4. Choice A has a sign error when adding: it computes x² - 4x + 1 + 3x - 5 as x² - 7x + 6, but -4x + 3x is -x, not -7x, and 1 - 5 is -4, not +6—double-check your like terms! The recipe for combining functions: (1) Write out what (f + g)(x) means: f(x) + g(x), (2) Substitute the actual function expressions, (3) Perform the operation carefully (watch signs!), (4) Simplify by combining like terms. Common pitfall: don't confuse (f + g)(x) with f(g(x)). The notation (f + g)(x) means 'add the functions,' while f(g(x)) means 'composition' (plug g into f). If you see +, -, ×, or ÷ in the notation, you're combining outputs, not composing functions!

Question 6

Let f(x)=xf(x)=\sqrt{x} and g(x)=1xg(x)=\dfrac{1}{x}. Find the domain of (f+g)(x)(f+g)(x).

  1. {x∣x≥0}\{x\mid x\ge 0\}
  2. {x∣x≠0}\{x\mid x\neq 0\}
  3. {x∣x>0}\{x\mid x>0\} (correct answer)
  4. {x∣x<0}\{x\mid x<0\}

Explanation: This question tests your ability to combine standard function types—like linear, quadratic, exponential, and constant functions—using arithmetic operations to build more complex functions. The domain of a combined function is where BOTH original functions are defined: if f needs x ≥ 0 and g needs x ≠ 2, then (f + g) needs both conditions: x ≥ 0 AND x ≠ 2. For division (f/g), we also need g(x) ≠ 0. It's the intersection of all the requirements! Looking at the domain of each function: f(x) = √x requires x ≥ 0, and g(x) = 1/x requires x ≠ 0. For (f + g)(x), we need BOTH conditions to be met: x ≥ 0 AND x ≠ 0, so x > 0 (since x = 0 would make g undefined). Combining all restrictions: domain is x > 0. Choice C correctly identifies the domain by intersecting the restrictions, giving {x | x > 0}. Choice A gives the correct domain for f alone, but misses that the combined function needs to satisfy BOTH domains: we need x ≥ 0 from f AND x ≠ 0 from g, so the domain is x > 0, not just x ≥ 0. For finding domains of combinations: make a list of ALL restrictions: (1) domain restrictions from f, (2) domain restrictions from g, (3) for (f/g), also where g(x) = 0. The domain of the combination is where ALL of these are satisfied simultaneously—the intersection of all conditions!

Question 7

Let f(x)=x2−4x+1f(x)=x^2-4x+1 and g(x)=3x−5g(x)=3x-5. Find (f+g)(x)(f+g)(x).​

  1. (f+g)(x)=x2−4x+1+(3x−5)(f+g)(x)=x^2-4x+1+(3x-5)
  2. (f+g)(x)=x2+12x−5(f+g)(x)=x^2+12x-5
  3. (f+g)(x)=x2−7x+6(f+g)(x)=x^2-7x+6
  4. (f+g)(x)=x2−x−4(f+g)(x)=x^2-x-4 (correct answer)

Explanation: This question tests your ability to combine standard function types—like linear, quadratic, exponential, and constant functions—using arithmetic operations to build more complex functions. When we combine functions using addition, we write (f + g)(x) = f(x) + g(x), which means 'add the outputs': for any input x, evaluate both f(x) and g(x), then add those results. For example, if f(x) = 2x + 1 and g(x) = x², then (f + g)(x) = (2x + 1) + x² = x² + 2x + 1. The result is a new function that combines both behaviors! To find (f + g)(x) where f(x) = x² - 4x + 1 and g(x) = 3x - 5, we add the expressions: (f + g)(x) = (x² - 4x + 1) + (3x - 5) = x² - 4x + 3x + 1 - 5. Simplifying: x² - x - 4, a new function that combines quadratic and linear behaviors into another quadratic. Choice B correctly combines the functions by adding and simplifying to x² - x - 4. Choice A has a sign error when adding: it computes x² - 4x + 1 + 3x - 5 as x² - 7x + 6, but -4x + 3x is -x, not -7x, and 1 - 5 is -4, not +6—double-check your like terms! The recipe for combining functions: (1) Write out what (f + g)(x) means: f(x) + g(x), (2) Substitute the actual function expressions, (3) Perform the operation carefully (watch signs!), (4) Simplify by combining like terms. Common pitfall: don't confuse (f + g)(x) with f(g(x)). The notation (f + g)(x) means 'add the functions,' while f(g(x)) means 'composition' (plug g into f). If you see +, -, ×, or ÷ in the notation, you're combining outputs, not composing functions!

Question 8

Let f(x)=xf(x)=\sqrt{x} and g(x)=1xg(x)=\dfrac{1}{x}. Find the domain of (f+g)(x)(f+g)(x).​

  1. {x∣x<0}\{x\mid x<0\}
  2. {x∣x≥0}\{x\mid x\ge 0\}
  3. {x∣x≠0}\{x\mid x\neq 0\}
  4. {x∣x>0}\{x\mid x>0\} (correct answer)

Explanation: This question tests your ability to combine standard function types—like linear, quadratic, exponential, and constant functions—using arithmetic operations to build more complex functions. The domain of a combined function is where BOTH original functions are defined: if f needs x ≥ 0 and g needs x ≠ 2, then (f + g) needs both conditions: x ≥ 0 AND x ≠ 2. For division (f/g), we also need g(x) ≠ 0. It's the intersection of all the requirements! Looking at the domain of each function: f(x) = √x requires x ≥ 0, and g(x) = 1/x requires x ≠ 0. For (f + g)(x), we need BOTH conditions to be met: x ≥ 0 AND x ≠ 0, so x > 0 (since x = 0 would make g undefined). Combining all restrictions: domain is x > 0. Choice C correctly identifies the domain by intersecting the restrictions, giving {x | x > 0}. Choice A gives the correct domain for f alone, but misses that the combined function needs to satisfy BOTH domains: we need x ≥ 0 from f AND x ≠ 0 from g, so the domain is x > 0, not just x ≥ 0. For finding domains of combinations: make a list of ALL restrictions: (1) domain restrictions from f, (2) domain restrictions from g, (3) for (f/g), also where g(x) = 0. The domain of the combination is where ALL of these are satisfied simultaneously—the intersection of all conditions!

Question 9

A material's stretch factor is modeled by multiplying a constant calibration factor and an exponential response. Let f(x)=3f(x)=3 and g(x)=2xg(x)=2^x. Find (f⋅g)(x)(f\cdot g)(x).​

  1. (f⋅g)(x)=6x(f\cdot g)(x)=6^x
  2. (f⋅g)(x)=3⋅2x(f\cdot g)(x)=3\cdot 2^x (correct answer)
  3. (f⋅g)(x)=23x(f\cdot g)(x)=2^{3x}
  4. (f⋅g)(x)=3+2x(f\cdot g)(x)=3+2^x

Explanation: This question tests your ability to combine standard function types—like linear, quadratic, exponential, and constant functions—using arithmetic operations to build more complex functions. When we combine functions using multiplication, we write (f · g)(x) = f(x) · g(x), which means 'multiply the outputs': for any input x, evaluate both f(x) and g(x), then multiply those results. For example, if f(x) = 2x and g(x) = x + 1, then (f · g)(x) = 2x · (x + 1) = 2x² + 2x. The result is a new function that scales one by the other! To find (f · g)(x) where f(x) = 3 and g(x) = 2^x, we multiply the expressions: (f · g)(x) = 3 · 2^x, which is already simplified as a scaled exponential function. This new function combines the constant calibration with the exponential response to model the stretch factor. Choice B correctly combines the functions by multiplying the constant and exponential, giving (f · g)(x) = 3 · 2^x. Choice D confuses multiplication with addition: the question asks for product, not sum. Make sure you're performing the operation that's actually requested! The recipe for combining functions: (1) Write out what (f · g)(x) means: f(x) · g(x), (2) Substitute the actual function expressions, (3) Perform the operation carefully, (4) Simplify by combining like terms. Real-world combinations make intuitive sense: stretch factor = calibration × response (multiply), which matches how we naturally combine quantities!

Question 10

Let f(x)=1xf(x)=\dfrac{1}{x} and g(x)=x2+1g(x)=x^2+1. Find the domain of (f+g)(x)(f+g)(x).

  1. {x∣x≠0}\{x\mid x\ne 0\} (correct answer)
  2. {x∣x≥0}\{x\mid x\ge 0\}
  3. {x∣x>0}\{x\mid x>0\}
  4. (−∞,∞)(-\infty,\infty)

Explanation: This question tests your ability to combine standard function types—like linear, quadratic, exponential, and constant functions—using arithmetic operations to build more complex functions. The domain of a combined function is where BOTH original functions are defined: if f needs x ≥ 0 and g needs x ≠ 2, then (f + g) needs both conditions: x ≥ 0 AND x ≠ 2. For division (f/g), we also need g(x) ≠ 0. It's the intersection of all the requirements! Looking at the domain of each function: f(x) = 1/x requires x ≠ 0, and g(x) = x² + 1 requires all real x (no restrictions). For (f + g)(x), we need BOTH conditions to be met: x ≠ 0 AND all real x, so just x ≠ 0. Choice B correctly identifies the domain as {x | x ≠ 0}. Choice C gives the domain for g alone, but misses that the combined function needs to satisfy BOTH domains: we need x ≠ 0 from f AND all reals from g, so the domain is {x | x ≠ 0}, not all reals. For finding domains of combinations: make a list of ALL restrictions: (1) domain restrictions from f, (2) domain restrictions from g, (3) for (f/g), also where g(x) = 0. The domain of the combination is where ALL of these are satisfied simultaneously—the intersection of all conditions!