Problem 17 - A Graphical Approach to Precalculus with Limits

Question

Use the intermediate value theorem to show that each function has a real zero between the two numbers given. Then, use a calculator to approximate the zero to the nearest hundredth. $$P(x)=x^{5}-2 x^{3}+1 ;-1.6 \text { and }-1.5$$

Step-by-Step Solution

Answer

The real zero is approximately at \( x = -1.56 \).

Step 1: Evaluate Function at Endpoints

Calculate the values of the function at the given points: First, evaluate at \( x = -1.6 \):\[P(-1.6) = (-1.6)^5 - 2(-1.6)^3 + 1 = -10.48576 + 8.192 + 1 = -1.29376.\]Next, evaluate at \( x = -1.5 \):\[P(-1.5) = (-1.5)^5 - 2(-1.5)^3 + 1 = -7.59375 + 6.75 + 1 = 0.15625.\]

Step 2: Apply Intermediate Value Theorem

Check for a sign change between \( P(-1.6) \) and \( P(-1.5) \): Since \( P(-1.6) = -1.29376 \) and \( P(-1.5) = 0.15625 \), there is a sign change from negative to positive, indicating a zero in the interval \([-1.6, -1.5]\). According to the Intermediate Value Theorem, because \( P(x) \) is continuous in the interval, a real zero exists between \(-1.6\) and \(-1.5\).

Step 3: Find Approximate Zero Using Calculator

Use a calculator to find the zero within the interval to the nearest hundredth.Approximate the zero by checking values between \(-1.6\) and \(-1.5\), refining the interval based on the sign changes.For example, by checking values closer: Evaluate at \( x = -1.56 \): \[P(-1.56) = (-1.56)^5 - 2(-1.56)^3 + 1 = -0.04976.\]Since \( P(-1.56) \) is close to zero, continue this process until the zero is refined to approximately \(-1.56\).

Key Concepts

Real Zero

A real zero of a function is a point where the function intersects the x-axis. In simple terms, it's the value of x that makes the function equal to zero. For the function given in the exercise, this means finding a value between -1.6 and -1.5 where the polynomial function becomes zero. To locate real zeros, we often utilize mathematical theorems and tools that help us pinpoint this intersection. Finding a real zero involves:

• Identifying a sign change in function values across an interval
• Using approximation methods to narrow down the x-value to a more precise figure.

Since polynomials are smooth and continuous over the real numbers, they will cross the x-axis when there's a change in sign. For example, as shown in the exercise, the sign changed from negative to positive between -1.6 and -1.5, indicating a real zero lies within this interval.

Polynomial Function

A polynomial function is an algebraic expression composed of variables (x), coefficients, and operations of addition, subtraction, and multiplication. These functions are characterized by their degree, which is determined by the highest power of x in the expression. In our exercise, the given polynomial is: \[ P(x) = x^5 - 2x^3 + 1 \]This is a fifth-degree polynomial because the highest power of x present is 5. Polynomial functions are incredibly important in mathematics due to their properties and behaviors, including their continuous nature, which is crucial for applying the Intermediate Value Theorem.Key features of polynomial functions include:

• Smooth and continuous curves with no breaks
• Relative ease of differentiation and integration
• A finite number of turning points

These features make them convenient for both mathematical analysis and practical applications, including the task given in the exercise.

Continuous Function

A continuous function is a function with no breaks, jumps, or holes in its graph. It can be drawn without lifting the pen from the paper. This smoothness is what allows for certain mathematical theorems, like the Intermediate Value Theorem, to hold true in the context of polynomial functions.For our polynomial \( P(x) = x^5 - 2x^3 + 1 \), continuity means:

• There are no sudden jumps in the values of the function as x changes
• The curve of the polynomial is unbroken over any interval
• The Intermediate Value Theorem can be applied safely

The Intermediate Value Theorem specifically states that for any value between \( P(-1.6) \) and \( P(-1.5) \), there must be at least one corresponding x-value. Thus, because \( P(x) \) is continuous over the interval [-1.6, -1.5] and there is a sign change, we are guaranteed a real zero within this range. This is an essential step in verifying the presence of a real zero when using the theorem in practice.