Problem 23 - Precalculus

Question

In Exercises \(21-45,\) find and simplify the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for the given function. $$ f(x)=6 $$

Step-by-Step Solution

Answer

The difference quotient simplifies to 0.

Step 1: Understand the Function

The function given is a constant function: \( f(x) = 6 \). This means that for any value of \( x \), the output is always 6.

Step 2: Substitute into the Difference Quotient Formula

The difference quotient formula is \( \frac{f(x+h) - f(x)}{h} \). Substitute \( f(x) = 6 \) and \( f(x+h) = 6 \) into the formula.

Step 3: Calculate the Numerator

Substitute the values we determined: \( f(x+h) = 6 \) and \( f(x) = 6 \) into the numerator of the difference quotient. We get: \( f(x+h) - f(x) = 6 - 6 = 0 \).

Step 4: Simplify the Difference Quotient

Substitute the simplified numerator back into the difference quotient: \( \frac{0}{h} \). Since the numerator is zero, the entire expression simplifies to \( 0 \).

Step 5: Final Conclusion

Thus, the difference quotient \( \frac{f(x+h)-f(x)}{h} \) simplifies to 0 for the function \( f(x) = 6 \).

Key Concepts

Constant Function

A constant function is one of the simplest types of functions you'll encounter in mathematics. It is defined as a function that always returns the same value, no matter what input you provide. In this case, the function is defined as \( f(x) = 6 \). This means that whether \( x \) is 1, 10, or even -50, the output remains 6.

Constant functions are important because they help us understand changes, or in this case, a lack of change. With the output always the same, there is no rise or slope when you graph this function. It's a flat line on a coordinate plane parallel to the x-axis, illustrating its unchanging nature across all values of \( x \).

Simplification

Simplification is the process of reducing an expression to its simplest form. When dealing with a difference quotient for a constant function like \( f(x) = 6 \), simplification becomes straightforward.

Initially, the difference quotient is given by the formula \( \frac{f(x+h) - f(x)}{h} \). In the case of a constant function, both \( f(x+h) \) and \( f(x) \) evaluate to 6.

So, substituting these values in, the expression \( f(x+h) - f(x) \) quickly reduces to \( 6 - 6 = 0 \).

Thus, you end up with \( \frac{0}{h} \). Regardless of the value of \( h \), as long as it isn't zero, dividing zero by \( h \) results in zero. Hence, the final simplified form of the difference quotient is 0.

Precalculus

Precalculus is often seen as the bridge between algebra and calculus. It introduces concepts that are foundational for understanding calculus, such as functions, their behaviors, and transformations. A key concept in precalculus, which is relevant here, is the difference quotient.

This quotient is a formula used to find the average rate of change of a function. It's akin to finding the slope of the secant line between two points on a graph of a function. However, with constant functions like \( f(x) = 6 \), no change or slope exists, which is why the difference quotient simplifies to zero.

Understanding these concepts in precalculus can make the transition to calculus smoother, where these ideas are expanded upon, especially when dealing with instantaneous rate of change—one of calculus's central themes.

Function Evaluation

Function evaluation involves determining the output of a function for a specific input. It's a fundamental concept that helps in understanding how functions behave.

When the constant function \( f(x) = 6 \) is evaluated, it simply returns 6 regardless of what number is used for \( x \). This means that for the difference quotient calculation, both \( f(x) \) and \( f(x+h) \) yield the same result of 6.

This illustrates that function evaluation repeatedly confirms the constant nature of \( f(x) \). In other scenarios, evaluating functions can involve more complex algebra, especially with polynomial or rational functions. However, with certain functions like the one discussed, the process is quite straightforward and underscores the constancy of the output in such situations.