Question

Find all the points of local maxima and local minima of the function: \[ f(x) = -\frac{3}{4}x^4 - 8x^3 - \frac{45}{2}x^2 + 105. \]

Hint:

To determine the nature of critical points, use \( f'(x) = 0 \) to find them and \( f''(x) \) to classify them as maxima, minima, or points of inflection.

Answer 1

Step 1: Compute the first derivative \( f'(x) \): \[ f'(x) = \frac{d}{dx}\left(-\frac{3}{4}x^4 - 8x^3 - \frac{45}{2}x^2 + 105\right) = -3x^3 - 24x^2 - 45x. \] 
Step 2: Factorize \( f'(x) \): \[ f'(x) = -3x(x^2 + 8x + 15) = -3x(x + 3)(x + 5). \] 
Step 3: Solve \( f'(x) = 0 \): \[ x = 0, \quad x = -3, \quad x = -5. \] These are the critical points where the slope of the tangent is zero. 
Step 4: Compute the second derivative \( f''(x) \): \[ f''(x) = \frac{d}{dx}(-3x^3 - 24x^2 - 45x) = -9x^2 - 48x - 45. \] 
Step 5: Use the second derivative test to determine the nature of each critical point: At \( x = -5 \): \[ f''(-5) = -9(-5)^2 - 48(-5) - 45 = -225 + 240 - 45 = -30 \quad ({negative, local maximum}). \] At \( x = -3 \): \[ f''(-3) = -9(-3)^2 - 48(-3) - 45 = -81 + 144 - 45 = 18 \quad ({positive, local minimum}). \] At \( x = 0 \): \[ f''(0) = -9(0)^2 - 48(0) - 45 = -45 \quad ({negative, local maximum}). \] 
Step 6: Calculate the function values at critical points: \[ f(-5) = -\frac{3}{4}(-5)^4 - 8(-5)^3 - \frac{45}{2}(-5)^2 + 105 = -468.75, \] \[ f(-3) = -\frac{3}{4}(-3)^4 - 8(-3)^3 - \frac{45}{2}(-3)^2 + 105 = 157.5, \] \[ f(0) = -\frac{3}{4}(0)^4 - 8(0)^3 - \frac{45}{2}(0)^2 + 105 = 105. \] 
Step 7: Conclusion: Local maxima at \( x = -5 \) (\( f(-5) = -468.75 \)) and \( x = 0 \) (\( f(0) = 105 \)). Local minima at \( x = -3 \) (\( f(-3) = 157.5 \)).