Problem 91 - Calculus: Early Transcendentals

Question

Applying the Intermediate Value Theorem Use the Intermediate Value Theorem to verify that the following equations have three solutions on the given interval. Use a graphing utility to find the approximate roots. $$70 x^{3}-87 x^{2}+32 x-3=0 ;(0,1)$$

Step-by-Step Solution

Answer

Based on the given equation and the analysis using the Intermediate Value Theorem (IVT), we can conclude that there are three solutions in the interval (0,1). The approximate roots obtained using a graphing utility are $$x \approx 0.123, x \approx 0.406, \text{ and } x \approx 0.786$$.

Step 1: Define the function

We are given the equation $$70x^3 - 87x^2 + 32x - 3 = 0$$ Define a function $$f(x) = 70x^3 - 87x^2 + 32x - 3$$

Step 2: Verify the conditions required for the IVT to apply

The IVT states that if a function $$f(x)$$ is continuous in a closed interval $$[a, b]$$, and $$k$$ is a value between $$f(a)$$ and $$f(b)$$, then there exists a value $$c$$ in the interval $$(a, b)$$ such that $$f(c) = k$$. Since polynomials are continuous functions in their domain, and the given interval (0,1) is a sub-interval of the domain of the given polynomial, the IVT can be applied.

Step 3: Check for intermediate values to find the intervals where the solutions could be

To check for intermediate values, evaluate the function at the endpoints of the given interval (0,1): $$f(0) = 70(0)^3 - 87(0)^2 + 32(0) - 3 = -3$$ $$f(1) = 70(1)^3 - 87(1)^2 + 32(1) - 3 = 70 - 87 + 32 - 3 = 12$$ Since $$f(0) < 0$$ and $$f(1) > 0$$, there is at least one value $$c$$ in the interval $$(0, 1)$$ for which $$f(c) = 0$$. To verify the existence of three solutions, we will check for local maxima and minima in the interval, which are potential locations for more solutions. The first derivative of the function is: $$f'(x) = 3 \cdot 70x^2 - 2 \cdot 87x + 32 = 210x^2 - 174x + 32$$ Find critical points by setting $$f'(x) = 0$$: $$210x^2 - 174x + 32 = 0$$ Factoring this equation is difficult, so we will resort to a graphing utility to find the critical points.

Step 4: Use a graphing utility to find the approximate roots

With the help of a graphing utility, we find the critical points to be approximately $$x \approx 0.273$$ and $$x \approx 0.727$$. By evaluating the function $$f(x)$$ at these critical points, we can check whether the function changes signs at each of these points: $$f(0.273) \approx -0.177$$ $$f(0.727) \approx 0.221$$ Since the function changes sign at these critical points, there must be three solutions in the interval (0,1). We can use the graphing utility to obtain the approximate roots of the equation. The graphing utility gives us the following approximate solutions: $$x \approx 0.123, x \approx 0.406, \text{ and } x \approx 0.786$$

Key Concepts

Polynomial Equations

Polynomial equations like the ones in our problem are mathematical statements involving sums of powers of a variable multiplied by coefficients. The general form of a polynomial equation of degree n is:

• \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0 \)

In the given exercise, we have a cubic polynomial because the highest power of the variable \( x \) is 3:

• \( 70x^3 - 87x^2 + 32x - 3 = 0 \)

Each polynomial equation can have as many real roots as its degree; in this case, three roots are possible. To solve these kinds of equations, we often look for values of \( x \) that satisfy the equation, making the polynomial equal to zero.

Continuity

Continuity is a key concept when discussing functions like polynomials. A function is continuous if, roughly speaking, you can draw it without lifting your pen off the paper. In mathematical terms, a function \( f(x) \) is continuous over an interval if there are no breaks, jumps, or holes in the curve of the function over that interval.

For the Intermediate Value Theorem to apply, the function must be continuous over the given interval. Fortunately, polynomial functions are continuous over the entire set of real numbers. Thus, our function \( f(x) = 70x^3 - 87x^2 + 32x - 3 \) is continuous over the interval \((0,1)\). This allows us to apply the theorem confidently, knowing there are no disruptions in continuity.

Graphing Utility

A graphing utility is a very helpful tool when solving complex polynomial equations. It allows us to visualize the behavior of a function, pinpoint approximate values for roots, and check points of interest like critical points where the function changes direction.

In our exercise, a graphing utility is used to:

• Find the critical points of the function.
• See where the function crosses the x-axis (the roots).
• Verify that the function changes signs, confirming the presence of roots.

By inputting the function \( f(x) = 70x^3 - 87x^2 + 32x - 3 \), the graph shows where solutions exist, and helps in estimating the roots as \( x \approx 0.123, x \approx 0.406, \text{ and } x \approx 0.786 \). Using technology reduces manual calculation errors and speeds up problem solving.

Critical Points

Critical points of a function occur where the derivative equals zero or is undefined. They help us identify where a function might change direction from increasing to decreasing (or vice versa). This is crucial in understanding the intervals where the function might cross the x-axis between points, indicating potential roots.

For the given problem, the first derivative \( f'(x) = 210x^2 - 174x + 32 \) gives us critical points when set to zero. Solving this using a graphing utility, we find:

• Critical points at \( x \approx 0.273 \) and \( x \approx 0.727 \).

By evaluating the function at these points and observing the changes in sign, we can establish that the function indeed passes through the x-axis, reaffirming there are three solutions in the interval \((0,1)\). This insight is crucial to utilizing the Intermediate Value Theorem effectively.