Math Problem Statement

If f is a linear function satisfying f(x + f(x)) = x + f(x), then f(5) can be equal to:

Solution

We are given that ff is a linear function and satisfies the functional equation:

f(x+f(x))=x+f(x)f(x + f(x)) = x + f(x)

We are tasked with determining f(5)f(5).

Step 1: Analyze the functional equation

Let’s first rewrite the given equation for clarity:

f(x+f(x))=x+f(x)f(x + f(x)) = x + f(x)

Since this holds for any xx, we can consider special cases or deduce properties of the function ff. Let’s assume f(x)f(x) is a linear function, which means we can write it in the form:

f(x)=ax+bf(x) = ax + b

where aa and bb are constants. We will substitute this expression into the functional equation and solve for aa and bb.

Step 2: Substitution into the functional equation

Substituting f(x)=ax+bf(x) = ax + b into the equation f(x+f(x))=x+f(x)f(x + f(x)) = x + f(x):

f(x+f(x))=f(x+ax+b)=a(x+ax+b)+b=ax+a2x+ab+bf(x + f(x)) = f(x + ax + b) = a(x + ax + b) + b = ax + a^2x + ab + b

On the right-hand side, we have:

x+f(x)=x+ax+bx + f(x) = x + ax + b

Now equate both sides:

ax+a2x+ab+b=x+ax+bax + a^2x + ab + b = x + ax + b

Step 3: Simplifying

We can simplify this equation by collecting like terms:

a2x+(ab+b)=xa^2x + (ab + b) = x

Now, comparing the coefficients of xx and the constant terms:

  • Coefficient of xx: a2=1a^2 = 1, so a=±1a = \pm 1.
  • Constant term: ab+b=0ab + b = 0, which simplifies to b(a+1)=0b(a + 1) = 0.

Step 4: Solve for aa and bb

  1. If a=1a = 1, the equation b(1+1)=0b(1 + 1) = 0 gives b=0b = 0.

    Thus, f(x)=xf(x) = x.

  2. If a=1a = -1, the equation b(1+1)=0b(-1 + 1) = 0 is automatically satisfied for any bb. Therefore, f(x)=x+bf(x) = -x + b.

Step 5: Determine f(5)f(5)

We now have two possible forms for f(x)f(x):

  1. If f(x)=xf(x) = x, then f(5)=5f(5) = 5.
  2. If f(x)=x+bf(x) = -x + b, we need additional information about bb to determine f(5)f(5). Since the problem doesn’t provide more constraints, both forms are potential solutions, but f(x)=xf(x) = x is the simplest and most natural solution under the assumption of linearity.

Conclusion

Thus, f(5)f(5) can be equal to 55.


Would you like a detailed exploration of any part of this solution? Here are some further questions you could explore:

  1. What would happen if the function was not linear?
  2. How does the assumption of linearity simplify the functional equation?
  3. Can we derive any additional properties of the function based on this result?
  4. What if we impose other conditions on the function ff?
  5. How would the solution change if ff were quadratic or of a higher degree?

Tip: Always verify assumptions like linearity when solving functional equations, as they can significantly simplify the process!

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Math Problem Analysis

Mathematical Concepts

Algebra
Functional Equations
Linear Functions

Formulas

f(x) = ax + b
f(x + f(x)) = x + f(x)

Theorems

Properties of linear functions

Suitable Grade Level

Grades 11-12