Questions & Answers: "Maxima and Minima"

To find the maximum value, we first find the derivative f'(x) = cos x - sin x. Setting f'(x) = 0 gives cos x = sin x, which implies tan x = 1, so x = π/4 within the given interval. The second derivative f''(x) = -sin x - cos x is negative at x = π/4 (f''(π/4) = -1/√2 - 1/√2 = -√2 < 0), confirming a local maximum. The value at x = π/4 is f(π/4) = sin(π/4) + cos(π/4) = 1/√2 + 1/√2 = 2/√2 = √2. Comparing this with the boundary values f(0) = 1 and f(π/2) = 1, the absolute maximum is √2. Option A is incorrect because 1 is the value at the boundaries, not the peak. Option C and D are incorrect based on the calculated trigonometric sum.

First, find the derivative: f'(x) = 3x² - 3. Set f'(x) = 0 to find critical points: 3(x² - 1) = 0, so x = 1 or x = -1. Now use the second derivative test: f''(x) = 6x. For x = 1, f''(1) = 6 > 0, indicating a local minimum. The value at x = 1 is f(1) = 1³ - 3(1) + 2 = 0. For x = -1, f''(-1) = -6 < 0, indicating a local maximum. The value at x = -1 is f(-1) = (-1)³ - 3(-1) + 2 = -1 + 3 + 2 = 4. Therefore, the local minimum value is 0. Option B represents the local maximum. Options C and D are not values at the critical points.

We are given x + y = 10, so y = 10 - x. The product function is P(x) = x(10 - x) = 10x - x². To maximize P, we find its derivative: P'(x) = 10 - 2x. Setting P'(x) = 0 gives x = 5. The second derivative P''(x) = -2 is negative, confirming that x = 5 yields a maximum. When x = 5, y = 10 - 5 = 5. The maximum product is P = 5 * 5 = 25. Option A (20) and B (24) are products of other pairs (like 28 or 46) which are less than 25. Option D (100) is the square of the sum, not the maximum product under the constraint.

To find local maxima/minima, we first find f'(x) = 6x² - 30x + 36. Setting f'(x) = 0 gives x² - 5x + 6 = 0, so (x-2)(x-3) = 0, giving critical points x=2 and x=3. Using the second derivative test, f''(x) = 12x - 30. At x=2, f''(2) = 24 - 30 = -6 (negative), indicating a local maximum. At x=3, f''(3) = 36 - 30 = 6 (positive), indicating a local minimum. Thus, the maximum occurs at x=2.

f'(x) = 3x², so f'(0) = 0. However, f''(x) = 6x, so f''(0) = 0, and the second derivative test is inconclusive. Using the first derivative test, f'(x) = 3x² is positive for x 0. Since the sign of the derivative does not change, there is no extremum. Because f''(x) changes sign at x=0 (from negative for x0), it is a point of inflection.

To find local extrema, we look for f'(x) = 0. For f(x) = e^x, f'(x) = e^x. However, e^x is always greater than 0 for all real values of x and never equals zero. Since there are no critical points where the derivative is zero or undefined, the function has no local maxima or minima.

For a quadratic function ax² + bx + c where a > 0, the minimum occurs at x = -b/2a. Here, x = -(-4)/(2*1) = 2. Substituting x=2 into the function: f(2) = (2)² - 4(2) + 7 = 4 - 8 + 7 = 3. Alternatively, completing the square gives f(x) = (x-2)² + 3, where the minimum value is clearly 3 when the squared term is zero.

f'(x) = 1 - 1/x². Setting f'(x) = 0 gives 1 = 1/x², so x² = 1. Since x > 0, we take x = 1. f''(x) = 2/x³, so f''(1) = 2 (positive), which confirms a minimum at x=1. The minimum value is f(1) = 1 + 1/1 = 2. This can also be solved using the AM-GM inequality: (x + 1/x)/2 ≥ √(x * 1/x) => x + 1/x ≥ 2.

Velocity v = ds/dt = 12t - 3t². To find the maximum velocity, we differentiate v with respect to t: dv/dt = 12 - 6t. Setting dv/dt = 0 gives 6t = 12, so t = 2. To confirm it's a maximum, d²v/dt² = -6 (negative). Thus, the velocity is maximum at t = 2 units of time.

Let sides be x and y. Perimeter P = 2(x+y), so y = (P/2) - x. Area A = xy = x(P/2 - x) = (Px/2) - x². To maximize area, dA/dx = P/2 - 2x = 0, which gives x = P/4. Since x = P/4, then y = P/2 - P/4 = P/4. Because x = y, the rectangle is a square.

According to the Second Derivative Test, for a function to have a local minimum at a point c, the first derivative must be zero (f'(c)=0, a stationary point) and the rate of change of the slope must be positive (f''(c)>0), meaning the curve is concave up. Option 1 describes a local maximum. Option 4 is inconclusive and requires further testing.

First, find critical points: f'(x) = 3x² - 3 = 0 => x² = 1 => x = 1, -1. Evaluate f(x) at critical points and endpoints: f(-2) = (-8) - 3(-2) + 3 = 1; f(-1) = (-1) - 3(-1) + 3 = 5; f(1) = 1 - 3(1) + 3 = 1; f(2) = 8 - 3(2) + 3 = 5. The maximum value reached in this set is 5.

Differentiating f(x), we get f'(x) = cos x - sin x. Setting f'(x) = 0 gives cos x = sin x, or tan x = 1, which means x = π/4 in the given interval. The value at x = π/4 is sin(π/4) + cos(π/4) = 1/√2 + 1/√2 = 2/√2 = √2. Checking endpoints: f(0) = 1 and f(π/2) = 1. Since √2 ≈ 1.414 > 1, the maximum value is √2.