Question
For certain real numbers a and b, the function f satisfies f(x) = ax + b + cx^2 and f(bx + a) = x + cx^2 for all real numbers x. Find a + b.
Answers
GPT-4o mini
We start with two functions defined by the given conditions:
1. \( f(x) = ax + b + cx^2 \)
2. \( f(bx + a) = x + cx^2 \)
First, we substitute \( bx + a \) into the function \( f(x) \):
\[
f(bx + a) = a(bx + a) + b + c(bx + a)^2
\]
Expanding the terms, we first compute \( a(bx + a) \):
\[
a(bx + a) = abx + a^2
\]
Next, we compute \( (bx + a)^2 \):
\[
(bx + a)^2 = b^2x^2 + 2abx + a^2
\]
Then, multiplying by \( c \) gives:
\[
c(bx + a)^2 = cb^2x^2 + 2cabx + ca^2
\]
Now, substituting into our expression for \( f(bx + a) \):
\[
f(bx + a) = abx + a^2 + b + cb^2x^2 + 2cabx + ca^2
\]
Collecting like terms for \(x\) and \(x^2\):
\[
f(bx + a) = (ab + 2cab)x + (a^2 + b + ca^2) + cb^2x^2
\]
We are given that
\[
f(bx + a) = x + cx^2
\]
This means we can equate the coefficients from both expressions. For the coefficient of \( x \):
\[
ab + 2cab = 1 \tag{1}
\]
For the constant term:
\[
a^2 + b + ca^2 = 0 \tag{2}
\]
For the coefficient of \( x^2 \):
\[
cb^2 = c \tag{3}
\]
Now we analyze equation (3). If \( c \neq 0 \), we can divide both sides by \( c \):
\[
b^2 = 1
\]
Thus, \( b = 1 \) or \( b = -1 \).
If \( c = 0 \), from equation (2), \( a^2 + b = 0 \), thus \( b = -a^2 \). We will explore the two cases for \( b \).
**Case 1: \( b = 1 \)**
Substituting \( b = 1 \) into (1):
\[
a(1) + 2ca = 1 \Rightarrow a + 2ca = 1 \tag{4}
\]
Substituting \( b = 1 \) into (2):
\[
a^2 + 1 + ca^2 = 0 \Rightarrow (1 + c)a^2 + 1 = 0
\]
Since this must hold for all \( a \), we find:
\[
1 + c = 0 \Rightarrow c = -1
\]
Substituting \( c = -1 \) back into (4):
\[
a - 2a = 1 \Rightarrow -a = 1 \Rightarrow a = -1
\]
Now we have \( a = -1 \) and \( b = 1 \).
Calculating \( a + b \):
\[
a + b = -1 + 1 = 0
\]
**Case 2: \( b = -1 \)**
Substituting \( b = -1 \) into (1):
\[
-a + 2ca = 1 \Rightarrow -a + 2ca = 1 \tag{5}
\]
Substituting \( b = -1 \) into (2):
\[
a^2 - 1 + ca^2 = 0 \Rightarrow (1 + c)a^2 - 1 = 0
\]
Setting the term with \( a^2 \) equal:
If \( 1 + c = 0 \) (as in the previous case), then:
\[
c = -1 \Rightarrow 0 = 1, \text{ no solution}.
\]
If \( 1 + c \neq 0 \), then rearranging gives:
\[
a^2 = \frac{1}{1 + c}
\]
Going back to (5) with \( c \) unknown doesn’t lead us to a new \( a \) that would satisfy all constraints. Hence, trying possible values yields complex or unsatisfactory solutions for this case.
Therefore, the only valid solution is from Case 1.
Final results yield:
\[
\boxed{0}
\]
1. \( f(x) = ax + b + cx^2 \)
2. \( f(bx + a) = x + cx^2 \)
First, we substitute \( bx + a \) into the function \( f(x) \):
\[
f(bx + a) = a(bx + a) + b + c(bx + a)^2
\]
Expanding the terms, we first compute \( a(bx + a) \):
\[
a(bx + a) = abx + a^2
\]
Next, we compute \( (bx + a)^2 \):
\[
(bx + a)^2 = b^2x^2 + 2abx + a^2
\]
Then, multiplying by \( c \) gives:
\[
c(bx + a)^2 = cb^2x^2 + 2cabx + ca^2
\]
Now, substituting into our expression for \( f(bx + a) \):
\[
f(bx + a) = abx + a^2 + b + cb^2x^2 + 2cabx + ca^2
\]
Collecting like terms for \(x\) and \(x^2\):
\[
f(bx + a) = (ab + 2cab)x + (a^2 + b + ca^2) + cb^2x^2
\]
We are given that
\[
f(bx + a) = x + cx^2
\]
This means we can equate the coefficients from both expressions. For the coefficient of \( x \):
\[
ab + 2cab = 1 \tag{1}
\]
For the constant term:
\[
a^2 + b + ca^2 = 0 \tag{2}
\]
For the coefficient of \( x^2 \):
\[
cb^2 = c \tag{3}
\]
Now we analyze equation (3). If \( c \neq 0 \), we can divide both sides by \( c \):
\[
b^2 = 1
\]
Thus, \( b = 1 \) or \( b = -1 \).
If \( c = 0 \), from equation (2), \( a^2 + b = 0 \), thus \( b = -a^2 \). We will explore the two cases for \( b \).
**Case 1: \( b = 1 \)**
Substituting \( b = 1 \) into (1):
\[
a(1) + 2ca = 1 \Rightarrow a + 2ca = 1 \tag{4}
\]
Substituting \( b = 1 \) into (2):
\[
a^2 + 1 + ca^2 = 0 \Rightarrow (1 + c)a^2 + 1 = 0
\]
Since this must hold for all \( a \), we find:
\[
1 + c = 0 \Rightarrow c = -1
\]
Substituting \( c = -1 \) back into (4):
\[
a - 2a = 1 \Rightarrow -a = 1 \Rightarrow a = -1
\]
Now we have \( a = -1 \) and \( b = 1 \).
Calculating \( a + b \):
\[
a + b = -1 + 1 = 0
\]
**Case 2: \( b = -1 \)**
Substituting \( b = -1 \) into (1):
\[
-a + 2ca = 1 \Rightarrow -a + 2ca = 1 \tag{5}
\]
Substituting \( b = -1 \) into (2):
\[
a^2 - 1 + ca^2 = 0 \Rightarrow (1 + c)a^2 - 1 = 0
\]
Setting the term with \( a^2 \) equal:
If \( 1 + c = 0 \) (as in the previous case), then:
\[
c = -1 \Rightarrow 0 = 1, \text{ no solution}.
\]
If \( 1 + c \neq 0 \), then rearranging gives:
\[
a^2 = \frac{1}{1 + c}
\]
Going back to (5) with \( c \) unknown doesn’t lead us to a new \( a \) that would satisfy all constraints. Hence, trying possible values yields complex or unsatisfactory solutions for this case.
Therefore, the only valid solution is from Case 1.
Final results yield:
\[
\boxed{0}
\]