Finding coefficients of a 3rd degree polynomial given the coefficient of the single integer, form of polynomial, and one value.

[NeeleshGupta023]

Hey people,
This is a question from the UIL Mathematics 2018 Invitational A, and I have absolutely no clue on how to solve it. Some help would be greatly appreciated. Thanks.

Let f(x) = ax^3 - bx + 3 where a and b are integers. If f(2) = -4, then f(-2) = ?

 

[Deleted member 4993]

Hey people,
This is a question from the UIL Mathematics 2018 Invitational A, and I have absolutely no clue on how to solve it. Some help would be greatly appreciated. Thanks.

Let f(x) = ax^3 - bx + 3 where a and b are integers. If f(2) = -4, then f(-2) = ?

Please refer to our guidelines of posting in this forum enunciated at: https://www.freemathhelp.com/forum/threads/guidelines-summary.109845/

Notice that f(2) = 8*a - 2*b + 3

f(-2) = -8*a + 2*b + 3

Please share your work/thoughts with us.

 

[ksdhart2]

How familiar are you with function notation? The notation used here, \(f(2)\), means we take the function \(f(x)\) and evaluate it at the point x = 2. To do so, we'll take the given expression for \(f(x)\) and replace every instance of \(x\) with 2:

\(f({\color{red}2}) = a({\color{red}2})^3 - b({\color{red}2}) + 3 = 8a - 2b + 3\)

What do you get when you use this same logic to evaluate \(f(-2)\)? This expression should look very similar to the previous one! Where does that lead you?

 

[ksdhart2]

Actually, having had a second think about this problem, I now believe there must be a typo somewhere, as the exercise exactly as written is impossible. We're told that \(a\) and \(b\) are integers, but we also know that

\(\displaystyle 8a - 2b + 3 = -4 \implies b = \frac{8a + 7}{2}\)

This relationship makes it impossible for \(a\) and \(b\) to both be integers. In fact, it will always be impossible for there to be integer solutions if f(2) is an even number:

\(\displaystyle 8a - 2b + 3 = 2k \implies b = 4a + \frac{3}{2} - k\)

Due to the definition of "even number," we know \(k\) must be an integer and we've stipulated that \(a\) is an integer as well, which means that \(b\) can never be an integer.

 

[NeeleshGupta023]

Actually, having had a second think about this problem, I now believe there must be a typo somewhere, as the exercise exactly as written is impossible. We're told that \(a\) and \(b\) are integers, but we also know that

\(\displaystyle 8a - 2b + 3 = -4 \implies b = \frac{8a + 7}{2}\)

This relationship makes it impossible for \(a\) and \(b\) to both be integers. In fact, it will always be impossible for there to be integer solutions if f(2) is an even number:

\(\displaystyle 8a - 2b + 3 = 2k \implies b = 4a + \frac{3}{2} - k\)

Due to the definition of "even number," we know \(k\) must be an integer and we've stipulated that \(a\) is an integer as well, which means that \(b\) can never be an integer.

I was thinking this exact same thing, and I worked out numbers for the equation 8a - 2b = -7, but I was not able to deduce any number combinations in which -7 could be reached. In this problem, there are 5 answer choices, but I thought there would be a way to attempt to find the equation for f(x) to evaluate at the point of negative 2. After these insightful responses, I am thinking that this question is just asking to find a number that could be in the given parameters of a third degree polynomial with a negative b that is shifted 3 points up.
The 5 given are answer choices are 4, 7, 10, -3, and -4.

 

[Deleted member 4993]

I was thinking this exact same thing, and I worked out numbers for the equation 8a - 2b = -7, but I was not able to deduce any number combinations in which -7 could be reached. In this problem, there are 5 answer choices, but I thought there would be a way to attempt to find the equation for f(x) to evaluate at the point of negative 2. After these insightful responses, I am thinking that this question is just asking to find a number that could be in the given parameters of a third degree polynomial with a negative b that is shifted 3 points up.
The 5 given are answer choices are 4, 7, 10, -3, and -4.

I do not know what did you mean by that point. The "answer" could be derived easily. If f(2) = -4, then f(-2) = ? :

f(x) = a*x³ - b*x +3

f(2) = -4 → -4 = a*(2)^3 - b*(2) +3 → 8a - 2b = -7

f(-2) = a*(-2)^3 - b*(-2) +3 → f(-2) =-8a + 2b + 3 = -(8a - 2b) + 3 = -(-7) + 3 = 7 + 3 = 10 ← answer

However, as KSD pointed out, 'a' and 'b' could not integers.