Question

Using the Intermediate Value Theorem In Exercises \(95-98\) , verify that the Intermediate Value Theorem applies to the indicated interval and find the value of \(c\) guaranteed by the theorem. $$ f(x)=x^{2}+x-1, \quad[0,5], \quad f(c)=11 $$

Short answer

Yes, the Intermediate Value Theorem applies to the function \(f(x)=x^2+x-1\) on the interval [0,5], and the value guaranteed by the theorem is \(c=3\).

Step-by-step solution

Verify the Conditions for Applying IVT

Start by checking if the function is continuous in the given interval \([0,5]\). The function \(f(x)=x^2+x-1\) is a polynomial, hence it is continuous on \([0,5]\). Then evaluate the function at the interval's endpoints. Calculate \(f(0)\) and \(f(5)\). It is easy to see that \(f(0) = -1\) and \(f(5) = 29\). Since \(f(0) < 11 < f(5)\), the conditions for the IVT are met. This means there must be a value \(c\) in the interval \([0,5]\) such that \(f(c)=11\).

Find the Value of c

Find the value of \(c\) for which \(f(c)=11\). This is done by setting \(f(x) =11\), and solving the resulting equation \(x^2 + x - 1 = 11\) or \(x^2 + x - 12 = 0\). The roots of this equation can be found using the quadratic formula \(x=(-b \pm \sqrt{b^2-4ac}) / 2a\), where \(a=1\), \(b=1\), and \(c=-12\). This yields \(x=3\) and \(x=-4\). Since only \(x=3\) lies in the interval \([0,5]\), we can conclude that \(c=3\).

Validation

Verify by substituting \(x=c=3\) into the function to ensure \(f(c)=11\). This yields \(f(3) = 3^2 + 3 - 1 = 9 + 3 - 1 = 11\), which confirms our solution.

Key concepts

Continuity of Polynomials

A critical property of polynomials, such as the function \(f(x) = x^2 + x - 1\), is their continuity. Continuity means that the function does not "jump" or "break" at any point in its domain. For polynomials, this property holds for all real numbers, including any given intervals. In the exercise, we have the interval \[0, 5\]. Polynomials maintain this smooth and uninterrupted behavior throughout this interval. This continuity allows the Intermediate Value Theorem (IVT) to apply, which is essential for solving problems where we need to find specific function values within an interval.

Quadratic Formula

The quadratic formula is a universal tool used to find the roots of quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is given by:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Let's apply this to the equation from our exercise, \(x^2 + x - 12 = 0\), where \(a = 1\), \(b = 1\), and \(c = -12\). Using the quadratic formula, we substitute the values to get:

• \(x = \frac{-1 \pm \sqrt{1 + 48}}{2}\)
• \(x = \frac{-1 \pm 7}{2}\)

This results in potential solutions: \(x = 3\) and \(x = -4\). Only \(x = 3\) fits in our specified interval \[0, 5\].

Solution of Polynomial Equations

Finding the value of \(c\) in the given polynomial equation is fundamental. In polynomial equations like this, solving involves expressing the equation in standard form, such as \(x^2 + x - 12 = 0\) for our case. This standard form allows us to apply methods such as factoring and the quadratic formula to find the roots. The roots are the solutions to the equation where the polynomial equals zero. In the current problem, solving yields the roots \(x = 3\) and \(x = -4\). Evaluating these roots within a specified interval helps us find the required value of \(c\), which is \(3\) in this case.

Interval Evaluation

When working with polynomial functions in a specific interval, evaluating the endpoints gives critical insight. For the function \(f(x) = x^2 + x - 1\) and the interval \[0, 5\], we start by calculating the function values at the boundaries:

• \(f(0) = -1\)
• \(f(5) = 29\)

The Intermediate Value Theorem tells us there’s a number \(c\) within this interval where \(f(c) = 11\), because \(11\) is between \(-1\) and \(29\). Knowing this, we search for \(c\) by solving the equation \(x^2 + x - 12 = 0\) and checking which solution(s) fit within our interval.

Function Roots

Roots of a function are the values that make the function equal to zero. For quadratic functions like \(f(x) = x^2 + x - 12 = 0\), roots can be found using various algebraic techniques, including factoring and the quadratic formula. In this problem, the quadratic formula provides the precise values of \(x = 3\) and \(x = -4\). These values indicate where the function crosses or touches the x-axis. The solution helps in identifying where intermediate values occur, aiding in validating the Intermediate Value Theorem within a specific interval. In this case, \(x = 3\) is the relevant value within our interval, giving us \(f(3) = 11\).