Problem 9

Question

Use the intermediate value theorem to show that each function has a real zero between the two numbers given. Then, use your calculator to approximate the zero to the nearest hundredth. $$P(x)=3 x^{2}-2 x-6 ; \quad 1 \text { and } 2$$

Step-by-Step Solution

Verified
Answer
There is a real zero between 1 and 2, approximately at 1.53.
1Step 1: Understand the Intermediate Value Theorem
The Intermediate Value Theorem states that if a continuous function \( f(x) \) takes values \( f(a) \) and \( f(b) \) at any two points \( a \) and \( b \), and if \( f(a) \) and \( f(b) \) have opposite signs, then there exists at least one number \( c \) between \( a \) and \( b \) such that \( f(c) = 0 \). This means we need to check the signs of \( P(1) \) and \( P(2) \).
2Step 2: Evaluate P(x) at the Given Points
Calculate \( P(1) = 3(1)^2 - 2(1) - 6 = 3 - 2 - 6 = -5 \). Next, calculate \( P(2) = 3(2)^2 - 2(2) - 6 = 12 - 4 - 6 = 2 \). Therefore, \( P(1) = -5 \) and \( P(2) = 2 \).
3Step 3: Apply the Intermediate Value Theorem
Since \( P(1) = -5 \) and \( P(2) = 2 \), and they have opposite signs, the Intermediate Value Theorem confirms that there is a real zero between \( x = 1 \) and \( x = 2 \).
4Step 4: Approximate the Zero Using a Calculator
Use a calculator to find the zero of \( P(x) = 3x^2 - 2x - 6 \) between 1 and 2. By using numerical methods, such as the bisection method or a calculator's root function, you find that the zero is approximately \( x \approx 1.53 \).

Key Concepts

Real Zeros of PolynomialsEvaluating FunctionsNumerical Approximation Methods
Real Zeros of Polynomials
Polynomials are mathematical expressions involving variables raised to whole-number exponents. The "real zeros" of a polynomial are solutions to the equation derived by setting the polynomial equal to zero. These are the points where the graph of the polynomial crosses or touches the x-axis. In simpler terms, they are the x-values for which the polynomial results in a value of zero. Finding real zeros can help us understand the behavior of polynomial functions.
  • The polynomial in this exercise is given as: \[ P(x) = 3x^2 - 2x - 6 \].
  • To find the real zeros, we want to find the values of \( x \) where \( P(x) = 0 \).
An important aspect to remember is that a polynomial of degree \( n\) can have up to \( n \) real zeros. In this case, because our polynomial is quadratic (degree 2), it can have up to 2 real zeros.
The Intermediate Value Theorem helps us identify at least one real zero within a given interval by examining the signs of the function's values at the interval's endpoints.
Evaluating Functions
Evaluating a function simply means finding the value of the function for a particular input. In this exercise, we evaluated the polynomial function \( P(x) = 3x^2 - 2x - 6 \) at different points.
  • First, we calculated \( P(1) = 3(1)^2 - 2(1) - 6 = -5 \).
  • Next, we figured out \( P(2) = 3(2)^2 - 2(2) - 6 = 2 \).
Why is this useful? Evaluating functions allows us to determine if a function changes signs over an interval, which is crucial for applying the Intermediate Value Theorem. Since \( P(1) \) and \( P(2) \) have opposite signs, it means that the function crosses the x-axis between \( x = 1 \) and \( x = 2 \). Thus, confirming the existence of a real zero within this interval. This step is foundational in accurately pinpointing where zero of the polynomial occurs.
Numerical Approximation Methods
At times, precisely calculating the zero of a polynomial by hand is challenging or impossible. This is where numerical approximation methods are beneficial. These methods help us estimate the root of the polynomial with the help of technology, like calculators or computers.
One popular method is the bisection method, which narrows down the interval where the zero resides by repeatedly halving the interval and checking the sign changes.
  • This process is repeated until the interval is sufficiently small to give a good approximation.
For our polynomial, using such methods led us to estimate that the real zero lies at \( x \approx 1.53 \). This value is found using a calculator's root function, which can apply methods like Newton's method or the bisection method internally. Knowing these numerical methods is essential, especially for higher-degree polynomials or complex functions where exact solutions are difficult to find manually.