- Identify intercepts: \((0,0)\) and \((4,0)\).
- Determine monotonicity: Decreasing on \((-\infty, 1)\), increasing on \((1, 4)\) and \((4, \infty)\).
- Identify extrema: Relative minimum at \((1, -\frac{27}{7})\).
- Determine concavity: Concave up on \((-\infty, 2)\) and \((4, \infty)\), concave down on \((2, 4)\); inflection points at \((2, -\frac{16}{7})\) and \((4, 0)\).
Explanation
- Problem Analysis and Basic Features
The function is given by \(f(x) = \frac{1}{7}x(x-4)^3\). This is a polynomial of degree 4, which means it is continuous and differentiable over the entire domain \((-\infty, \infty)\).
- Intercepts:
- The x-intercepts occur when \(f(x) = 0\), which gives \(x = 0\) or \(x - 4 = 0 \implies x = 4\). The points are \((0, 0)\) and \((4, 0)\).
- The y-intercept is found by calculating \(f(0)\), which is \(f(0) = \frac{1}{7}(0)(0-4)^3 = 0\). This confirms the origin \((0, 0)\) is an intercept.
- End Behavior:
- The leading term is \(\frac{1}{7}x^4\). As \(x \to \infty\), \(f(x) \to \infty\). As \(x \to -\infty\), \(f(x) \to \infty\). The graph opens upwards on both sides.
- First Derivative Analysis and Extrema
To find the critical points, we compute the first derivative \(f'(x)\):
\[f'(x) = \frac{1}{7} \left[ (1)(x-4)^3 + x \cdot 3(x-4)^2 \right] = \frac{1}{7}(x-4)^2(x-4 + 3x) = \frac{1}{7}(x-4)^2(4x-4) = \frac{4}{7}(x-1)(x-4)^2\]
Setting \(f'(x) = 0\) gives the critical points at \(x = 1\) and \(x = 4\).
- Intervals of Increase/Decrease:
- On \((-\infty, 1)\), test \(x=0\): \(f'(0) = \frac{4}{7}(-1)(-4)^2 = -\frac{64}{7} < 0\), so the function is decreasing.
- On \((1, 4)\), test \(x=2\): \(f'(2) = \frac{4}{7}(1)(-2)^2 = \frac{16}{7} > 0\), so the function is increasing.
- On \((4, \infty)\), test \(x=5\): \(f'(5) = \frac{4}{7}(4)(1)^2 = \frac{16}{7} > 0\), so the function is increasing.
- Relative Extrema:
- At \(x=1\), the function changes from decreasing to increasing, so there is a relative minimum. The value is \(f(1) = \frac{1}{7}(1)(-3)^3 = -\frac{27}{7} \approx -3.86\).
- At \(x=4\), the function does not change direction; it is a horizontal tangent point but not a relative extremum.
- Second Derivative and Concavity Analysis
We compute the second derivative \(f''(x)\):
\[f''(x) = \frac{d}{dx} \left[ \frac{4}{7}(x-1)(x-4)^2 \right] = \frac{4}{7} \left[ (1)(x-4)^2 + (x-1) \cdot 2(x-4) \right]\]
\[f''(x) = \frac{4}{7}(x-4)(x-4 + 2x - 2) = \frac{4}{7}(x-4)(3x-6) = \frac{12}{7}(x-4)(x-2)\]
Setting \(f''(x) = 0\) gives inflection points at \(x = 2\) and \(x = 4\).
- Concavity:
- On \((-\infty, 2)\), test \(x=0\): \(f''(0) = \frac{12}{7}(-4)(-2) = \frac{96}{7} > 0\), so the function is concave up.
- On \((2, 4)\), test \(x=3\): \(f''(3) = \frac{12}{7}(-1)(1) = -\frac{12}{7} < 0\), so the function is concave down.
- On \((4, \infty)\), test \(x=5\): \(f''(5) = \frac{12}{7}(1)(3) = \frac{36}{7} > 0\), so the function is concave up.
- Inflection Points:
- At \(x=2\), \(f(2) = \frac{1}{7}(2)(-2)^3 = -\frac{16}{7} \approx -2.29\).
- At \(x=4\), \(f(4) = 0\).
- Final Graphing Summary
The function \(f(x) = \frac{1}{7}x(x-4)^3\) has intercepts at \((0,0)\) and \((4,0)\). It has a relative minimum at \((1, -\frac{27}{7})\) and inflection points at \((2, -\frac{16}{7})\) and \((4, 0)\). The function decreases on \((-\infty, 1)\) and increases on \((1, 4)\) and \((4, \infty)\). It is concave up on \((-\infty, 2)\) and \((4, \infty)\), and concave down on \((2, 4)\). There are no vertical or horizontal asymptotes as it is a polynomial.
Examples
Engineers and architects often analyze functions like this to model the curvature of structural elements. By identifying points of inflection, where the concavity changes, they can determine the points of maximum stress or transitions in the shape of a beam, ensuring that the structure is both aesthetically pleasing and structurally sound. For example, in bridge design, understanding the inflection points of a curve allows for precise placement of supports to manage load distribution effectively.