Question

In Exercises \(5-8,\) find the two \(x\) -intercepts of the function \(f\) and show that \(f^{\prime}(x)=0\) at some point between the two \(x\) -intercepts. $$ f(x)=x^{2}-x-2 $$

Short answer

The two x-intercepts of the function \(f(x)=x^{2}-x-2\) are \(x = -1\) and \(x = 2\). The derivative of the function is \(f'(x) = 2x - 1\). The derivative equals zero at \(x = 0.5\), which lies between the two x-intercepts. So there exists a number \(c = 0.5\) between the x-intercepts for which the derivative \(f'(c)=0\).

Step-by-step solution

Find x-intercepts of the function

Setting the function \(f(x)=x^{2}-x-2\) equal to zero yields the equation \(x^{2}-x-2=0\). This equation can be solved with the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), and \(c = -2\). Solving this gives \(x_1 = -1\) and \(x_2 = 2\). Thus, the x-intercepts of the function are \(x = -1\) and \(x = 2\).

Find the derivative of the function

The derivative of the function \(f(x)=x^{2}-x-2\) can be found using the power rule for differentiation, which states that the derivative of \(x^n\) is \(n*x^{n-1}\). Applying this rule to each term of the function gives \(f'(x) = 2x - 1\).

Find a point where the derivative is 0

Setting the derivative \(f'(x) = 2x - 1\) equal to zero and solving for \(x\) yields \(x = 0.5\). Since the value 0.5 lies between the two x-intercepts -1 and 2, it is therefore demonstrated that there exists a number \(c = 0.5\) between the x-intercepts for which the derivative \(f'(c)=0\)

Key concepts

Quadratic Formula

The quadratic formula is a cornerstone concept when dealing with polynomial equations of the second degree, commonly known as quadratic functions. A quadratic function has the general form of f(x) = ax^2 + bx + c, where a, b, and c are constants, and a ≠ 0. The formula provides a method to find the roots, or x-intercepts, of any quadratic equation.

The quadratic formula is given by: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]
It instructs us to subtract and add the square root of the discriminant b^2 - 4ac to the opposite of b, all over twice the value of a. The discriminant can indicate the nature of the roots; when it's positive, there are two real and distinct roots, if zero, there is exactly one real root (known as a double root), and if negative, the roots are complex and not real.

Understanding the quadratic formula is essential as it provides a direct approach to find the x-intercepts for quadratic functions such as f(x) = x^2 - x - 2. Using the given constants a = 1, b = -1, and c = -2, the roots can be calculated, resulting in two x-intercepts that are crucial for graphing the function and understanding its behavior.

Derivative of Polynomial Functions

The process of differentiation is a fundamental operation in calculus, used to determine the derivative of a function, which represents the rate at which the function's value changes. For polynomial functions, the derivative at a point gives the slope of the tangent line to the graph of the function at that point. This concept becomes especially important when locating maxima and minima, or when analyzing the function's increasing and decreasing behavior.

Polynomial functions are sums of terms consisting of a coefficient multiplied by a variable to a power, such as f(x) = ax^n. The derivative of this polynomial function can be found by applying the power rule for differentiation.

In our example, f(x) = x^2 - x - 2 is a polynomial function. Taking the derivative term by term results in f'(x) = 2x - 1. This simplification is essential for further analysis of the function, such as determining the critical points where the derivative is zero, indicating potential local maxima or minima, or inflection points of the function's graph.

Power Rule for Differentiation

The power rule for differentiation is a straightforward yet powerful tool in calculus. It allows one to quickly find the derivative of a term with a power of x. According to the power rule, if you have a term of the form ax^n, where a is a coefficient and n is a positive integer, the derivative of that term is anx^(n-1).

Applying the Power Rule

For the function f(x) = x^2 - x - 2, we apply the power rule to each term individually:

• The derivative of x^2 is 2x, as indicated by bringing down the exponent and subtracting one from it.
• For the linear term -x, which is essentially x^1, the derivative is -1.
• And the constant -2 has a derivative of 0, as constants do not change.

Putting it all together gives us f'(x) = 2x - 1. By setting the derivative equal to zero, you can solve for x to find the critical points, which are often related to the function's extreme values and can provide insight into the behavior of the function between its intercepts.

The power rule's utility extends beyond simple calculations, playing a crucial role in various applications like finding velocity in physics, profit maximization in economics, and even in complex artificial intelligence algorithms where optimization is key.