Question

Let \(f(x)=x^2+ax+b\) and \(g(x)=f(x+1)-f(x-1)\). If \(f(x)≥0\) for all real \(x\), and \(g(20)=72\), then the smallest possible value of \(b\) is

• A. 1
• B. 16
• C. 0
• D. 4

Answer 1

The Correct Option is D

A function is provided:

\[ f(x) = x^2 + ax + b \]

A new function is defined:

\[ g(x) = f(x+1) - f(x-1) \]

Step 1: Expand \( f(x+1) \) and \( f(x-1) \)

\[ f(x+1) = (x+1)^2 + a(x+1) + b = x^2 + 2x + 1 + ax + a + b \] \[ f(x-1) = (x-1)^2 + a(x-1) + b = x^2 - 2x + 1 + ax - a + b \]

Step 2: Calculate \( g(x) \) by subtraction

\[ g(x) = f(x+1) - f(x-1) \] \[ = [x^2 + 2x + 1 + ax + a + b] - [x^2 - 2x + 1 + ax - a + b] \] \[ = 4x + 2a \]

Step 3: Apply the condition \( g(20) = 72 \)

\[ g(20) = 4(20) + 2a = 80 + 2a = 72 \] This implies \( 2a = -8 \), so \( a = -4 \).

Step 4: Rewrite \( f(x) \) with the determined value of \( a \)

\[ f(x) = x^2 - 4x + b \]

Completing the square, we get: \( f(x) = (x - 2)^2 + (b - 4) \).

Step 5: Determine the condition for \( f(x) \ge 0 \) for all \( x \)

The term \( (x - 2)^2 \) is always non-negative. The minimum value of \( f(x) \) occurs at \( x = 2 \), which is: \[ f(2) = (2 - 2)^2 + (b - 4) = b - 4 \]

For \( f(x) \ge 0 \) to hold true for all \( x \), the minimum value must be non-negative: \[ b - 4 \ge 0 \Rightarrow b \ge 4 \]

Final Answer:

The minimum possible value for \( b \) is \( \boxed{4} \).