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Question

Difficulty: HardFunction Evaluation, Composition, and Properties

Let the functions ff and gg be defined for all real numbers by f(x)=x2−6x+7f(x) = x^2 - 6x + 7 and g(x)=∣2x−5∣g(x) = |2x - 5|. What is the sum of all real values of xx for which f(g(x))=14f(g(x)) = 14?

  1. A
    3
  2. 5Answer
  3. C
    6
  4. D
    8
  5. E
    10

Answer

The sum of all possible real values of xx is 55.
The correct answer is 55. Setting f(g(x))=14f(g(x)) = 14 yields (g(x))2−6g(x)+7=14(g(x))^2 - 6g(x) + 7 = 14, which simplifies to (g(x))2−6g(x)−7=0(g(x))^2 - 6g(x) - 7 = 0. Factoring this quadratic gives g(x)=7g(x) = 7 or g(x)=−1g(x) = -1. Because g(x)=∣2x−5∣g(x) = |2x - 5| cannot be negative, we discard the negative case. Solving ∣2x−5∣=7|2x - 5| = 7 gives 2x−5=7⇒x=62x - 5 = 7 \Rightarrow x = 6 and 2x−5=−7⇒x=−12x - 5 = -7 \Rightarrow x = -1. The sum of these values is 6+(−1)=56 + (-1) = 5.

Step-by-Step Solution

1
Set up the composite equation f(g(x))=14f(g(x)) = 14 by substituting g(x)g(x) into f(x)f(x).
(g(x))2−6g(x)+7=14(g(x))^2 - 6g(x) + 7 = 14
This defines the relation between g(x)g(x) and the target output value.
2
Rearrange the equation into a standard quadratic form and solve for g(x)g(x).
(g(x))2−6g(x)−7=0⇒(g(x)−7)(g(x)+1)=0(g(x))^2 - 6g(x) - 7 = 0 \Rightarrow (g(x) - 7)(g(x) + 1) = 0, so g(x)=7g(x) = 7 or g(x)=−1g(x) = -1.
Factoring the quadratic equation gives the possible values for the inner function g(x)g(x).
3
Apply the definition of g(x)g(x) to solve for xx and discard invalid cases.
Since g(x)=∣2x−5∣g(x) = |2x - 5| must be non-negative, the case g(x)=−1g(x) = -1 has no real solutions. For the case g(x)=7g(x) = 7, we have ∣2x−5∣=7|2x - 5| = 7.
The range of an absolute value function is [0,∞)[0, \infty), making negative outputs impossible.
4
Solve the absolute value equation ∣2x−5∣=7|2x - 5| = 7 by setting up both positive and negative cases.
Either 2x−5=7⇒2x=12⇒x=62x - 5 = 7 \Rightarrow 2x = 12 \Rightarrow x = 6, or 2x−5=−7⇒2x=−2⇒x=−12x - 5 = -7 \Rightarrow 2x = -2 \Rightarrow x = -1.
An absolute value equation ∣u∣=c|u| = c splits into u=cu = c and u=−cu = -c.
5
Sum the valid solutions for xx.
6+(−1)=56 + (-1) = 5
The question asks for the sum of all real values of xx that satisfy the equation.

Key Concept

Solving equations involving composite functions and absolute values
Estimated Time:2m 0s
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Topics

Function Evaluation, Composition, and Properties

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