Question

Use a graphing utility to graph \(f\) and \(f^{\prime}\) on the interval \([-2,2]\). $$ f(x)=x(x+1) $$

Short answer

The function \(f(x) = x(x+1)\) is a parabola with roots at \(x=-1\) and \(x=0\), and the derivative of the function is \(f^{\prime}(x) = 2x+1\), which is graphed as a straight line on the interval \([-2,2]\).

Step-by-step solution

Function Analysis

Let's start with understanding the function \(f(x) = x(x+1)\). It's a quadratic function, which is a parabola that opens upward and has roots at \(x=-1\) and \(x=0\).

Derivation of the function

Using differentiation rules, let's find the derivative of the function. The derivative \(f^{\prime}(x)\) of \(f(x)\) is found using the product rule which states: \((u*v)^{\prime}= u^{\prime}*v+u*v^{\prime}\) where for \(f(x) = x(x+1)\), \(u = x\), and \(v = x+1\). Thus, \(f^{\prime}(x)= 1*(x+1)+x*1 = 2x+1\).

Graphing the function and its derivative

To graph this, select the interval \([-2,2]\) in the graphing tool. Then enter the original equation \(f(x) = x(x+1)\) and the derived equation \(f^{\prime}(x) = 2x+1\). The graph of \(f(x) = x(x+1)\) is a parabola and the graph of \(f^{\prime}(x)=2x+1\), that represents the slope of \(f(x)\), is a straight line.

Key concepts

Quadratic Functions

Quadratic functions are one of the fundamental types of algebraic expressions. They can be recognized by their characteristic form, which is:
\[ f(x) = ax^{2} + bx + c \]
where the variables represent constants and determine the shape and position of a parabola. When graphing a quadratic function, you will see a U-shaped curve known as a parabola.

The quadratic function in our exercise, \( f(x) = x(x + 1) \), simplifies to:
\[ f(x) = x^{2} + x \]
With a leading coefficient of 1 (since \( a = 1 \)), we can see that the graph will open upwards. The parabola will have its vertex at the minimum point since the coefficient of the squared term is positive.

Another key feature of quadratic functions are the roots or zeros, which are the x-values where the function crosses the x-axis. In our example, this happens where \( f(x) = 0 \) or at \( x = -1 \) and \( x = 0 \), giving us invaluable points for sketching the graph accurately.

Product Rule Differentiation

Differentiation is a process used to calculate the derivative of a function, which essentially provides us with the slope of the function at any given point. The product rule is a key differentiation tool used when you need to find the derivative of two multiplied functions.

The product rule states that the derivative of a product of two functions \( u(x) \) and \( v(x) \) is:
\[ (u \times v)' = u' \times v + u \times v' \]
In the given exercise, we have the function \( f(x) = x(x+1) \), which is a product of \( u(x) = x \) and \( v(x) = x+1 \). To find the derivative \( f'(x) \), we use the product rule:
\[ f'(x) = (x)'(x+1) + x(1) \] This simplifies to:
\[ f'(x) = x + 1 + x \] \[ f'(x) = 2x + 1 \]
The resulting derivative is a linear function, which means it will graph as a straight line, as opposed to the parabola of the original quadratic function.

Parabola Graphing

When it comes to graphing a parabola, understanding its vertex and roots is essential. The vertex is the highest or lowest point on the graph, and in a quadratic function such as \( f(x) = ax^{2} + bx + c \), it can be found using the formula:
\[ h = -\frac{b}{2a} \]
Once you know the vertex, plot it on the coordinate system along with the roots of the function to begin shaping your parabola.

For the function \( f(x) = x(x+1) \), which expands to \( f(x) = x^{2} + x \), the roots are located at \( x = -1 \) and \( x = 0 \). The vertex of this particular function is at \( h = -\frac{1}{2(1)} = -0.5 \), with the corresponding y-value found by substituting \( h \) back into the function.

Graphing utility programs can be very helpful for visualizing the parabola and its corresponding derivative. By using the interval \( [-2,2] \) as given in the exercise, students can observe not only the shape and location of the parabola but also how the derivative function, represented as a straight line, depicts the slope of the tangent to the parabola at any given x-value on the graph.