Math Problem Statement

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.

Solution

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Math Problem Analysis

Mathematical Concepts

Calculus
Critical Points
Derivatives
Quadratic Functions

Formulas

First derivative f'(x) = 2ax + b
Critical point x = -b/(2a)
Second derivative f''(x) = 2a

Theorems

Second Derivative Test
Critical Point Definition

Suitable Grade Level

Grades 11-12