Applications Of Derivatives – Maxima and Minima

Answer

Step 1: Find derivative: \( f'(x) = 2x - 6 \) Step 2: Set \( f'(x) > 0 \) for increasing: \( 2x - 6 > 0 \Rightarrow x > 3 \) Step 3: Set \( f'(x) < 0 \) for decreasing: \( 2x - 6 < 0 \Rightarrow x <…

Answer

Step 1: \( f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1) \) Step 2: Critical points: \( x = -1, 2 \) Step 3: Sign analysis: - For \( x < -1 \): \( f'(x) > 0 \Rightarrow \) increasing - F…

Answer

Step 1: Use quotient rule: \( f'(x) = \frac{(x^2 + 1)(1) - x(2x)}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} \) Step 2: Denominator always positive. Numerator: \( 1 - x^2 = (1 - x)(1 + x) \) Step 3:…

Answer

Step 1: \( f'(x) = -\sin x \) Step 2: For \( x \in (0, \pi) \), \( \sin x > 0 \Rightarrow -\sin x < 0 \) Step 3: So \( f'(x) < 0 \) in \( (0, \pi) \) Therefore, \( f(x) \) is decreasing in \( [0, \…

Answer

Step 1: \( f'(x) = 1 + \sin x \) Step 2: Since \( -1 \leq \sin x \leq 1 \), then: \( 1 + \sin x \geq 1 - 1 = 0 \) So \( f'(x) \geq 0 \) for all \( x \) Hence, \( f(x) \) is increasing everywhere.

Answer

Step 1: \( f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1) \) Step 2: Set \( f'(x) = 0 \Rightarrow x = -1, 3 \) Step 3: Sign change analysis: - At \( x = -1 \): \( f'(x) \) changes from + …

Answer

Step 1: \( f'(x) = 2x - 4 \) Step 2: Set \( f'(x) = 0 \Rightarrow x = 2 \) Step 3: Sign analysis: - For \( x < 2 \), \( f'(x) < 0 \) - For \( x > 2 \), \( f'(x) > 0 \) So \( f'(x) \) changes from −…

Answer

Step 1: \( f'(x) = 6x^2 - 6x - 12 = 6(x + 1)(x - 2) \) Step 2: Critical points: \( x = -1, 2 \) Step 3: Sign analysis: - At \( x = -1 \): \( f'(x) \) changes + to − → local max - At \( x = 2 \): \( …