- The critical points of the function \(f(x) = 6x^4 - 8x^3 - 87x^2 + 90x + 26\) are identified as \(x = -2.5\), \(x = 0.5\), and \(x = 3\).
- Based on the second derivative test, \(x = -2.5\) and \(x = 3\) are local minima, while \(x = 0.5\) is a local maximum.
- Comparing function values at these points and the endpoints \(-5\) and \(5\), the absolute maximum is \(\boxed{2151}\) and the absolute minimum is \(\boxed{-383.375}\).
Explanation
- Identify Critical Points
To find the critical points of the function \(f(x) = 6x^4 - 8x^3 - 87x^2 + 90x + 26\) on the interval \([-5, 5]\), we first find its derivative \(f'(x)\):
\[f'(x) = 24x^3 - 24x^2 - 174x + 90\]
We set the derivative equal to zero to find the critical points:
\[24x^3 - 24x^2 - 174x + 90 = 0\]
Dividing the entire equation by 6 simplifies it to:
\[4x^3 - 4x^2 - 29x + 15 = 0\]
Solving this cubic equation yields the critical points at \(x = -2.5\), \(x = 0.5\), and \(x = 3\). Since all these values fall within the interval \([-5, 5]\), they are all valid critical points.
- Determine Nature of Critical Points
To determine the nature of these critical points, we examine the second derivative \(f''(x)\):
\[f''(x) = 72x^2 - 48x - 174\]
We evaluate \(f''(x)\) at each critical point:
- For \(x = -2.5\): \(f''(-2.5) = 72(-2.5)^2 - 48(-2.5) - 174 = 450 + 120 - 174 = 396\). Since \(f''(-2.5) > 0\), \(x = -2.5\) is a local minimum.
- For \(x = 0.5\): \(f''(0.5) = 72(0.5)^2 - 48(0.5) - 174 = 18 - 24 - 174 = -180\). Since \(f''(0.5) < 0\), \(x = 0.5\) is a local maximum.
- For \(x = 3\): \(f''(3) = 72(3)^2 - 48(3) - 174 = 648 - 144 - 174 = 330\). Since \(f''(3) > 0\), \(x = 3\) is a local minimum.
- Find Absolute Extrema
To find the absolute extrema, we evaluate the function \(f(x)\) at the critical points and the endpoints of the interval \([-5, 5]\):
- At \(x = -5\): \(f(-5) = 6(-5)^4 - 8(-5)^3 - 87(-5)^2 + 90(-5) + 26 = 6(625) - 8(-125) - 87(25) - 450 + 26 = 3750 + 1000 - 2175 - 450 + 26 = 2151\).
- At \(x = -2.5\): \(f(-2.5) = 6(-2.5)^4 - 8(-2.5)^3 - 87(-2.5)^2 + 90(-2.5) + 26 = -383.375\).
- At \(x = 0.5\): \(f(0.5) = 6(0.5)^4 - 8(0.5)^3 - 87(0.5)^2 + 90(0.5) + 26 = 48.625\).
- At \(x = 3\): \(f(3) = 6(3)^4 - 8(3)^3 - 87(3)^2 + 90(3) + 26 = -217\).
- At \(x = 5\): \(f(5) = 6(5)^4 - 8(5)^3 - 87(5)^2 + 90(5) + 26 = 6(625) - 8(125) - 87(25) + 450 + 26 = 3750 - 1000 - 2175 + 450 + 26 = 1051\).
Comparing these values, the absolute maximum is 2151 (at \(x = -5\)) and the absolute minimum is -383.375 (at \(x = -2.5\)).
Examples
Imagine an entrepreneur analyzing the profit trend of a new product over its first 5 months. The function \(f(x)\) models the profit (in thousands of dollars) based on the month. Identifying critical points allows the owner to find when the profit hits its peak (local maximum) or its worst slump (local minimum). By checking the endpoints of the period, the owner ensures they know the absolute best and worst financial positions during the entire timeframe, supporting better long-term business planning.