Use calculus to prove that the relative minimum or maximum for any function​ f, as shown​ below, occurs at x equals negative StartFraction b Over 2 a EndFractionx=−
b
2a.
f left parenthesis x right parenthesis equals ax squared plus bx plus c comma a not equals 0f(x)=ax2+bx+c, a≠0 Determine the critical values of f left parenthesis x right parenthesis equals ax squared plus bx plus cf(x)=ax2+bx+c. A critical value can occur where f prime left parenthesis x right parenthesisf′(x)equals=0 and also where f prime left parenthesis x right parenthesisf′(x) does not exist. Since f prime left parenthesis x right parenthesisf′(x) is a​ polynomial, there are no​ x-values where f prime left parenthesis x right parenthesisf′(x) does not exist.​ Therefore, the only critical values will be where f prime left parenthesis x right parenthesisf′(x)equals=0. Solve the equation f prime left parenthesis x right parenthesisf′(x)equals=0 for x.

To prove that the relative minimum or maximum of the quadratic function \( f(x) = ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \), let's go through the steps using calculus.

### Step 1: Find the first derivative of \( f(x) \)
We need to find the critical points by first taking the derivative of the function \( f(x) = ax^2 + bx + c \).

\[
f'(x) = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b
\]

### Step 2: Set the derivative equal to zero
Critical points occur where the first derivative is equal to zero or where the derivative does not exist. Since \( f'(x) = 2ax + b \) is a polynomial, it exists for all values of \( x \). Thus, we only need to find where the derivative equals zero:

\[
2ax + b = 0
\]

### Step 3: Solve for \( x \)
Solve the equation \( 2ax + b = 0 \) for \( x \):

\[
2ax = -b
\]

\[
x = -\frac{b}{2a}
\]

This value of \( x \) represents the critical point.

### Step 4: Determine if it's a minimum or maximum
To confirm whether this critical point corresponds to a minimum or maximum, we take the second derivative of \( f(x) \).

\[
f''(x) = \frac{d}{dx}(2ax + b) = 2a
\]

- If \( a > 0 \), then \( f''(x) = 2a > 0 \), indicating the function is concave up at \( x = -\frac{b}{2a} \), meaning this point is a **relative minimum**.
- If \( a < 0 \), then \( f''(x) = 2a < 0 \), indicating the function is concave down at \( x = -\frac{b}{2a} \), meaning this point is a **relative maximum**.

### Conclusion:
The relative minimum or maximum of the quadratic function \( f(x) = ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \).

Would you like more details or have any questions?

Here are 5 related questions to expand this understanding:
1. How does the second derivative test confirm concavity and the type of critical point?
2. What happens if \( a = 0 \) in the function \( f(x) = ax^2 + bx + c \)?
3. Can you find the minimum or maximum value of \( f(x) \) at \( x = -\frac{b}{2a} \)?
4. How does this result relate to the vertex of a parabola in geometry?
5. Can we generalize this process for higher-degree polynomials?

**Tip:** The formula \( x = -\frac{b}{2a} \) also gives the x-coordinate of the vertex of a parabola.

Use calculus to prove that the relative minimum or maximum for any function f(x) = ax^2 + bx + c, a≠0 occurs at x = -b/(2a). Determine the critical values of f(x) = ax^2 + bx + c.

Learn how to use calculus to prove that the relative minimum or maximum of the quadratic function f(x) = ax^2 + bx + c occurs at x = -b/(2a). This problem covers key calculus concepts such as derivatives, critical points, and the second derivative test, suitable for students in Grades 11-12.

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