Problem 16 - Calculus

Question

Sketch the graph of \(f(x),\) and use this graph to sketch the graph of \(f^{\prime}(x)\). $$f(x)=x(x-1)$$

Step-by-Step Solution

Answer

Sketch \( f(x) \), a parabola, and \( f'(x) \), a line with slope 2.

Step 1: Find the Derivative of the Function

Before we sketch the graph of the derivative, we need to calculate it.Given, \( f(x) = x(x-1) \).Let's expand \( f(x) \): \( f(x) = x^2 - x \).Now, find the derivative \( f'(x) \):\[ f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(x) = 2x - 1 \].The derivative is \( f'(x) = 2x - 1 \).

Step 2: Sketch the Graph of the Original Function

To sketch \( f(x) = x(x-1) = x^2 - x \), identify key features:- **Intercepts**: Set \( f(x) = 0 \): \[ x(x-1) = 0 \] Roots are \( x = 0 \) and \( x = 1 \).- **Vertex**: Since \( f(x) \) is a parabola opening upwards, find the vertex by using \( x = \frac{-b}{2a} \) where \( a = 1, b = -1 \): \[ x = \frac{-(-1)}{2 \times 1} = \frac{1}{2} \].- **Value at Vertex**: \( f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} = -\frac{1}{4} \).- **Shape**: It's a U-shaped curve passing through the points \( (0, 0), (1, 0) \) and reaching a minimum at \( \left( \frac{1}{2}, -\frac{1}{4} \right) \).

Step 3: Use the Graph of f(x) to Sketch f'(x)

The graph of \( f'(x) = 2x - 1 \) is a straight line because it's linear.- **Intercept**: Set \( f'(x) = 0 \): \[ 2x - 1 = 0 \Rightarrow x = \frac{1}{2} \].- **Intercept of the line on y-axis**: When \( x = 0 \), \( f'(0) = -1 \).- **Slope**: The slope is 2, so for every 1 unit increase in \( x \), \( f'(x) \) increases by 2 units.- **Critical Points**: Derivative equals zero at \( x = \frac{1}{2} \), indicating a vertex point of the original function.The line crosses the y-axis at \( -1 \) and x-axis at \( x = \frac{1}{2} \). This indicates where the U-shaped parabola of \( f(x) \) has its minimum point and changes directions.

Key Concepts

Derivative

In calculus, the **derivative** represents the rate at which a quantity changes. For a function, the derivative at any given point gives the slope of the tangent line to the graph at that point. For the function \( f(x) = x^2 - x \), we found the derivative by applying basic differentiation rules to each term:

• The derivative of \( x^2 \) is \( 2x \).
• The derivative of \( x \) is \( 1 \).

Thus, the derivative of the entire function is \( f'(x) = 2x - 1 \).
This linear function indicates how steeply \( f(x) \) is climbing or falling at any point \( x \).
The role of the derivative is crucial in understanding the behavior of the original function, especially in determining where the function's graph slopes upwards, downwards or stays flat.

Parabola

A **parabola** is a specific type of graph that represents a quadratic function. It is defined by the equation \( ax^2 + bx + c \).
The standard shape of a parabola is a symmetric "U" curve.
In our function \( f(x) = x^2 - x \), the parabola opens upwards, which is characteristic of positive leading coefficients (\( a > 0 \)).Parabolas have several important features:

• Intercepts: These are points where the graph crosses the x-axis. For \( f(x) \), the intercepts are at \( x = 0 \) and \( x = 1 \).
• Direction: Determined by the sign of \( a \). Here, the positive \( a \) means it opens upwards.

The clear symmetry of a parabola makes it easier to sketch once these key characteristics are identified.

Vertex of a Parabola

The **vertex** of a parabola is its highest or lowest point. For an upward-opening parabola, the vertex is a minimum point.
To calculate the vertex for the function \( f(x) = x^2 - x \), we use the formula for the \( x \)-coordinate, \( x = \frac{-b}{2a} \).

• Here, \( a = 1 \) and \( b = -1 \), thus \( x = \frac{1}{2} \).
• The y-coordinate, found by substituting \( x = \frac{1}{2} \) into \( f(x) \), is \( f\left(\frac{1}{2}\right) = -\frac{1}{4} \).

Hence, the vertex is at \( \left( \frac{1}{2}, -\frac{1}{4} \right) \). This is where the function reaches its minimum value and changes direction.
Identifying the vertex is integral in graph sketching as it helps in drawing the curve accurately.

Slope of a Line

The **slope of a line** is a measure of its steepness and indicates how the y-value changes with the x-value. Mathematically, the slope is the ratio of the vertical change to the horizontal change between two points on the line.
For the derivative \( f'(x) = 2x - 1 \), the expression \( 2x - 1 \) is linear, meaning it's a straight line with a slope of 2.

• This slope tells us that for every increase of 1 unit in \( x \), \( f'(x) \) increases by 2 units.
• The line will cross the y-axis where \( x = 0 \), resulting in \( f'(0) = -1 \).

This concept helps in understanding the rate of change of the original function, \( f \), at any point.
The slope dictates the angle and direction of the derivative’s graph, which provides insights into where the original function is increasing or decreasing.