Suppose for all integers x, there are two function f and g such that $f(x) + f(x-1) - 1 = 0$ and $g(x) = x^2$. If $f(x^2 - x) = 5$, then the value of the sum $f(g(5)) + g(f(5))$ is

We break down this problem step by step, working with the given conditions to find the pattern in function $f$.

We have:

• $f(x) + f(x-1) - 1 = 0$ for all integers $x$
• $g(x) = x^2$
• $f(x^2 - x) = 5$

We need to find: $f(g(5)) + g(f(5))$

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From $f(x^2 - x) = 5$, we can substitute any integer value for $x$ to get specific information about $f$.

We try $x = 1$:

$f(1^2 - 1) = f(1 - 1) = f(0) = 5$

We now know $f(0) = 5$. This gives us a starting point!

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The condition $f(x) + f(x-1) - 1 = 0$ can be rewritten as:

$f(x) = 1 - f(x-1)$

This means if we know $f$ at any point, we can find $f$ at the next integer.

Starting with $f(0) = 5$:

For $x = 1$: $f(1) = 1 - f(0) = 1 - 5 = -4$

For $x = 2$: $f(2) = 1 - f(1) = 1 - (-4) = 5$

For $x = 3$: $f(3) = 1 - f(2) = 1 - 5 = -4$

For $x = 4$: $f(4) = 1 - f(3) = 1 - (-4) = 5$

For $x = 5$: $f(5) = 1 - f(4) = 1 - 5 = -4$

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We observe from our results:

• $f(0) = 5$
• $f(1) = -4$
• $f(2) = 5$
• $f(3) = -4$
• $f(4) = 5$
• $f(5) = -4$

The function $f$ alternates between two values!

• $f(\text{even integer}) = 5$
• $f(\text{odd integer}) = -4$

This pattern works because since $f(x) = 1 - f(x-1)$, the function keeps flipping between $1-5=-4$ and $1-(-4)=5$.

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Now we can find $f(g(5)) + g(f(5))$:

First, we find each part:

• $g(5) = 5^2 = 25$
• $f(g(5)) = f(25)$

Since 25 is odd: $f(25) = -4$

• $f(5) = -4$ (from our pattern, since 5 is odd)
• $g(f(5)) = g(-4) = (-4)^2 = 16$

Therefore: $f(g(5)) + g(f(5)) = f(25) + g(-4) = -4 + 16 = 12$

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Answer: 12

When we have a recurrence relation like $f(x) = 1 - f(x-1)$, we look for alternating patterns. This type of relation often creates functions that flip between two fixed values!