Problem 54 - Thomas Calculus

Question

A function value Show that the function \(F(x)=(x-a)^{2}\) . \((x-b)^{2}+x\) takes on the value \((a+b) / 2\) for some value of \(x .\)

Step-by-Step Solution

Answer

The function takes the value \(\frac{a+b}{2}\) at \(x = \frac{a+b}{2}\).

Step 1: Understanding the Problem

We need to show that the function \( F(x) = (x-a)^2 \cdot (x-b)^2 + x \) takes on the specific value \( \frac{a+b}{2} \) for some value of \( x \). Our goal is to find \( x \) such that \( F(x) = \frac{a+b}{2} \).

Step 2: Set the Equation

We set the equation \( F(x) = \frac{a+b}{2} \). Thus, we have \( (x-a)^2 \cdot (x-b)^2 + x = \frac{a+b}{2} \).

Step 3: Analyze the Function Structure

Observe that the term \( (x-a)^2 \cdot (x-b)^2 \) is a quartic polynomial, which suggests complexity. However, a simplification might exist based on the symmetry when \( x \) is around the midpoint of \( a \) and \( b \).

Step 4: Identify a Potential Solution Point

Consider the potential symmetry point \( x = \frac{a+b}{2} \). Substituting it into the function might highlight symmetric properties or solutions.

Step 5: Substitute the Potential Solution

Substitute \( x = \frac{a+b}{2} \) into the function:\[ F\left(\frac{a+b}{2}\right) = \left(\frac{a+b}{2}-a\right)^2 \cdot \left(\frac{a+b}{2}-b\right)^2 + \frac{a+b}{2} \]This simplifies to:\[ F\left(\frac{a+b}{2}\right) = \left(\frac{b-a}{2}\right)^2 \cdot \left(\frac{a-b}{2}\right)^2 + \frac{a+b}{2} \]Note that the squared terms make the polynomial expand uniformly around this midpoint.

Step 6: Simplify and Verify

The first term \( \left(\frac{b-a}{2}\right)^2 \cdot \left(\frac{a-b}{2}\right)^2 \) equals zero because each squared difference between the same terms negates, leading to zero. Thus:\[ F\left(\frac{a+b}{2}\right) = 0 + \frac{a+b}{2} = \frac{a+b}{2} \]This confirms that for \( x = \frac{a+b}{2} \), the function \( F(x) \) evaluates to the required value \( \frac{a+b}{2} \).

Step 7: Conclusion

Therefore, we've shown that when \( x = \frac{a+b}{2} \), the function \( F(x) = (x-a)^2 \cdot (x-b)^2 + x \) indeed takes on the value \( \frac{a+b}{2} \), satisfying the condition of the exercise.

Key Concepts

Quartic Polynomials

Quartic polynomials are expressions where the highest degree of the variable, typically denoted as "x," is four. In the function \( F(x) = (x-a)^2 \cdot (x-b)^2 + x \), the product \((x-a)^2 \cdot (x-b)^2\) is indeed a quartic polynomial because if you expand it, the highest power of \(x\) will be \(x^4\). This can make quartic functions complex due to their high degree, leading to multiple real or complex roots.

Here are some key points about quartic polynomials:

• They can have up to four real or complex roots.
• The graph of a quartic polynomial can have up to three turning points.
• Quartic functions can exhibit various shapes, depending on their coefficients.

Understanding quartic polynomials is important as they appear in numerous mathematical contexts, from geometry to physics. They can describe phenomena ranging from the motion seen in classical mechanics to the more abstract realms of algebraic geometry.

Symmetry in Functions

Symmetry in functions provides a tool to simplify and understand such functions better. Functions can exhibit a variety of symmetries, but in the context of our function \( F(x) = (x-a)^2 \cdot (x-b)^2 + x \), identifying symmetry helps to pinpoint a particular value of \( x \) that might satisfy a condition.

• One common type of symmetry is around the line \( x = h \), known as line symmetry.
• In the problem at hand, recognizing that \( x = \frac{a+b}{2} \) is a potential symmetry point helps in testing values efficiently.

By assessing whether substituting \( x = \frac{a+b}{2} \) leads the function to satisfy the desired conditions, we leverage symmetry to check potential solutions without exhaustive testing. Symmetry implies that changes around key points, like midpoints, might yield significant insights into the function's behavior.

Finding Function Values

Finding the value of a function for a specific input is a typical task in calculus. This involves substituting a particular value into the function and performing the necessary computations to determine the output. For our function \( F(x) = (x-a)^2 \cdot (x-b)^2 + x \), verifying that it takes on the value \( \frac{a+b}{2} \) when \( x = \frac{a+b}{2} \) involves a few steps:

• Substitute \( x = \frac{a+b}{2} \) into the function.
• Simplify the expression, often using algebraic identities or properties.
• Check if the simplified result matches the desired value.

In this case, after substitution and simplification, the squared terms cancel out, and what remains is the additional term from the function, \( \frac{a+b}{2} \). This reflects how particular values can lead to elegant solutions, emphasizing the importance of methodically substituting and simplifying in calculus to achieve desired results.