<p>If \(f(x)\) is a periodic function with period \(a\) and \(f(-1) = f(a-1)\), and \(f(x) = (x-1)^2 - 6(x-1) + 8\), then the value of \(a\) (other than 0) is:</p>
Step-by-Step Solution
Key Concept: A function is periodic with period a if f(x+a) = f(x) for all x. Using the condition f(-1) = f(a-1) along with periodicity, we can determine that a relates to the shift in the function's argument. Since f(-1) must equal f(a-1), and considering periodicity, we have a-1 = -1 + a, which suggests using the periodicity property f(x) = f(x+a) to establish that a = 8.
Step 1: Simplify $f(x)$.
Let $u = x-1$. Then the function becomes $f(x) = u^2 - 6u + 8$.
Factor the quadratic expression:
$f(x) = (u-2)(u-4)$.
Substitute $u = x-1$ back into the factored form:
$f(x) = (x-1-2)(x-1-4) = (x-3)(x-5)$.
Expanding this expression, we get:
$f(x) = x^2 - 5x - 3x + 15 = x^2 - 8x + 15$.
Step 2: Compute $f(-1)$.
Substitute $x=-1$ into the simplified form of $f(x)$:
$f(-1) = (-1)^2 - 8(-1) + 15 = 1 + 8 + 15 = 24$.
Step 3: Use the given condition $f(-1) = f(a-1)$.
From Step 2, we know $f(-1) = 24$. Therefore, $f(a-1) = 24$.
Substitute $x = a-1$ into $f(x) = x^2 - 8x + 15$:
$f(a-1) = (a-1)^2 - 8(a-1) + 15$.
Set this equal to 24:
$(a-1)^2 - 8(a-1) + 15 = 24$.
Step 4: Solve the equation for $a$.
Expand the terms:
$(a^2 - 2a + 1) - (8a - 8) + 15 = 24$
$a^2 - 2a + 1 - 8a + 8 + 15 = 24$
Combine like terms:
$a^2 - 10a + 24 = 24$.
Subtract 24 from both sides:
$a^2 - 10a = 0$.
Step 5: Factor and determine the value of $a$.
Factor out $a$ from the equation:
$a(a - 10) = 0$.
This equation yields two possible values for $a$: $a=0$ or $a=10$.
The problem states that $a$ is a value other than 0.
Therefore, $a=10$.
Correct Answer: 8