3. Find a linear function \( f(x)=a x+b \) given that: a) \( f(2 x+3) \equiv-2 \) b) \( f(1-x) \equiv 4 x+1 \) c) \( f(f(x)) \equiv 4 x+3 \)

To find the linear function \( f(x) = ax + b \), we can consider the given conditions one by one. **For condition a:** 1. We start with \( f(2x + 3) = a(2x + 3) + b \). 2. Simplifying, we get \( 2ax + (3a + b) \). 3. We know this is equal to -2, which involves no \( x \) terms, indicating that \( 2a = 0 \) implies \( a = 0 \) and thus \( 3a + b = -2 \) leads to \( b = -2 \). 4. Hence, from this condition, we find that \( f(x) \) is not of the typical linear form any further. **For condition b:** 1. We compute \( f(1 - x) = a(1 - x) + b = a - ax + b \). 2. Simplifying gives \( -ax + (a + b) \) which must equal \( 4x + 1 \). 3. This leads to two equations: \( -a = 4 \) and \( a + b = 1 \). 4. So we have \( a = -4 \) and substituting, \( -4 + b = 1 \) gives \( b = 5 \). 5. Therefore, we have the candidate \( f(x) = -4x + 5 \) from this condition. **For condition c:** 1. Now we can verify our function against the requirement \( f(f(x)) = f(-4x + 5) \). 2. Calculate \( f(-4x + 5) = -4(-4x + 5) + 5 = 16x - 20 + 5 = 16x - 15 \). 3. This must equal \( 4x + 3 \), which leads to a discrepancy as it’s not satisfied. Putting this process together, we find that checking each condition leads us towards \( f(x) = -4x + 5 \) being the best candidate while keeping in mind the contradictory nature of some setups.