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Difficulty: HardFunction Evaluation, Composition, and Properties

Let the functions ff and gg be defined by f(x)=∣2x−5∣f(x) = |2x - 5| and g(x)=x+7g(x) = \sqrt{x + 7} for all real numbers in their respective domains. If aa is a real number such that (f∘g)(a)=3(f \circ g)(a) = 3, what is the sum of all possible values of aa?

Answer: 3

Answer

The sum of all possible values of aa is 33.
The correct answer is 33. The composition (f∘g)(a)=3(f \circ g)(a) = 3 translates to f(g(a))=∣2a+7−5∣=3f(g(a)) = |2\sqrt{a + 7} - 5| = 3. This absolute value relation splits into two equations: 2a+7−5=32\sqrt{a + 7} - 5 = 3 and 2a+7−5=−32\sqrt{a + 7} - 5 = -3. Solving the first equation yields a+7=4⇒a=9\sqrt{a + 7} = 4 \Rightarrow a = 9. Solving the second equation yields a+7=1⇒a=−6\sqrt{a + 7} = 1 \Rightarrow a = -6. Since both 99 and −6-6 are greater than or equal to −7-7, they are within the domain of the radical function. The sum of these values is 9+(−6)=39 + (-6) = 3.

Step-by-Step Solution

1
Express the composition (f∘g)(a)(f \circ g)(a) in terms of aa.
∣2a+7−5∣=3|2\sqrt{a + 7} - 5| = 3
By definition of function composition, (f∘g)(a)=f(g(a))(f \circ g)(a) = f(g(a)). Substituting g(a)=a+7g(a) = \sqrt{a + 7} into the expression for f(x)f(x) yields f(g(a))=∣2g(a)−5∣=∣2a+7−5∣f(g(a)) = |2g(a) - 5| = |2\sqrt{a + 7} - 5|.
2
Set up the two algebraic cases to eliminate the absolute value.
2a+7−5=32\sqrt{a + 7} - 5 = 3 or 2a+7−5=−32\sqrt{a + 7} - 5 = -3
An absolute value equation of the form ∣u∣=c|u| = c (where c≥0c \geq 0) has two possible cases: u=cu = c or u=−cu = -c.
3
Solve the first equation case for aa.
a=9a = 9
Adding 55 to both sides of 2a+7−5=32\sqrt{a + 7} - 5 = 3 gives 2a+7=82\sqrt{a + 7} = 8. Dividing by 22 gives a+7=4\sqrt{a + 7} = 4. Squaring both sides yields a+7=16a + 7 = 16, which gives a=9a = 9.
4
Solve the second equation case for aa.
a=−6a = -6
Adding 55 to both sides of 2a+7−5=−32\sqrt{a + 7} - 5 = -3 gives 2a+7=22\sqrt{a + 7} = 2. Dividing by 22 gives a+7=1\sqrt{a + 7} = 1. Squaring both sides yields a+7=1a + 7 = 1, which gives a=−6a = -6.
5
Verify domain constraints and sum the valid solutions.
33
Both a=9a = 9 and a=−6a = -6 satisfy the domain requirement for g(x)=x+7g(x) = \sqrt{x+7}, which is x≥−7x \geq -7. The sum of these two valid values is 9+(−6)=39 + (-6) = 3.

Key Concept

Evaluating and solving equations containing composite functions, absolute values, and radical functions.

Alternative Method

Instead of expanding the composition immediately, substitute a temporary variable u=g(a)=a+7u = g(a) = \sqrt{a+7}. The equation becomes f(u)=3⇒∣2u−5∣=3f(u) = 3 \Rightarrow |2u - 5| = 3. Solve this simplified absolute value equation to get 2u−5=3⇒u=42u - 5 = 3 \Rightarrow u = 4, and 2u−5=−3⇒u=12u - 5 = -3 \Rightarrow u = 1. Next, substitute back a+7\sqrt{a+7} for uu: solving a+7=4\sqrt{a+7} = 4 yields a=9a = 9, and solving a+7=1\sqrt{a+7} = 1 yields a=−6a = -6. Summing these two solutions gives 9+(−6)=39 + (-6) = 3.
Estimated Time:2m 0s
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Topics

Function Evaluation, Composition, and Properties

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