Problem 23 - Calculus A New Horizon

Question

Use any method to find the relative extrema of the function \(f\). $$f(x)=x(x-1)^{2}$$

Step-by-Step Solution

Answer

The function has a local minimum at \(x = 1\) and a local maximum at \(x = \frac{1}{3}\).

Step 1: Identify the Function

The given function is \(f(x) = x(x-1)^{2}\). Our goal is to find the relative extrema (i.e., local maxima or minima).

Step 2: Find the First Derivative

To locate critical points, we first find the derivative \(f'(x)\) using the product rule. Let \(u = x\) and \(v = (x-1)^2\). Then the derivative of \(f(x)\) using the product rule \((uv)' = u'v + uv'\) is: \[f'(x) = 1(x-1)^2 + x\cdot 2(x-1)\cdot 1 \]Simplifying, we get: \[f'(x) = (x-1)^2 + 2x(x-1)\] Simplifying further, \[f'(x) = x^2 - 2x + 1 + 2x^2 - 2x \] \[f'(x) = 3x^2 - 4x + 1\]

Step 3: Set the Derivative to Zero

Set \(f'(x) = 0\) to find the critical points.\[3x^2 - 4x + 1 = 0 \]This is a quadratic equation which we can solve using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here \(a = 3\), \(b = -4\), \(c = 1\).

Step 4: Solve the Quadratic Equation

Using the quadratic formula: \[x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(1)}}{2(3)}\] Which simplifies to: \[x = \frac{4 \pm \sqrt{16 - 12}}{6} \] \[x = \frac{4 \pm 2}{6} \] Thus, solving gives:\[x = 1 \, \text{or} \, x = \frac{1}{3}\]

Step 5: Test Critical Points

To determine whether these points are maxima or minima, we can use the second derivative test. Find the second derivative \(f''(x)\): \[f''(x) = \frac{d}{dx}(3x^2 - 4x + 1)\] \[f''(x) = 6x - 4\] Evaluate \(f''(x)\) at \(x = 1\) and \(x = \frac{1}{3}\):- At \(x = 1\): \(f''(1) = 6(1) - 4 = 2\), which is positive, indicating a local minimum.- At \(x = \frac{1}{3}\): \(f''\left(\frac{1}{3}\right) = 6\left(\frac{1}{3}\right) - 4 = -2\), which is negative, indicating a local maximum.

Key Concepts

First Derivative

The first derivative of a function is a tool that helps us understand rates of change. For finding relative extrema, the first derivative is essential because it tells us where functions change their slope direction. In simpler terms, where the slope of the tangent to the curve transitions from positive to negative or vice versa. When calculated, the first derivative represents the slope of the original function.

To find it, we often use derivative rules like the product rule, chain rule, or power rule. In our example, we had the function \(f(x) = x(x-1)^2\). By using the product rule, we derived \(f'(x) = 3x^2 - 4x + 1\). This new expression tells us where \(f(x)\) might reach a peak or valley.

Critical Points

Critical points occur where the first derivative is zero or undefined. These points are like milestones along the journey of a function where something important happens. To find them, set the first derivative equal to zero and solve for \(x\).

In this exercise, we had \(3x^2 - 4x + 1 = 0\). Solving for this gives the values for critical points. These are the potential candidates for relative maxima or minima. They're crucial because maxima and minima find their roots in these points. Not all critical points will lead to an extremum, but examining them is vital for a deeper understanding of the function's behavior.

Second Derivative Test

The second derivative test helps us confirm the nature of each critical point found from the first derivative. The second derivative gives us information about the concavity of the function at a point. If \(f''(x) > 0\), the graph is concave up at \(x\), suggesting a local minimum. Conversely, if \(f''(x) < 0\), the graph is concave down, indicating a local maximum.

This test provides a systematic approach to distinguish between these points easily. In our scenario, \(f''(x) = 6x - 4\) was calculated. It helped us determine that at \(x = 1\), the function has a local minimum, while at \(x = \frac{1}{3}\), it has a local maximum.

Quadratic Formula

The quadratic formula is a powerful algebraic tool used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]It can reliably find the roots or solutions of any quadratic equation, providing us with critical points in our study of functions.

In our task, it helped solve \(3x^2 - 4x + 1 = 0\) and pinpointed \(x = 1\) and \(x = \frac{1}{3}\) as critical points. Each solution derived from this formula narrowed down the points where the function potentially has relative extrema. It's an efficient way to transition from algebraic expressions to more graphical insights.