Compute The Difference Quotient Answer As Much As Possible

Compute the difference quotient answer as much as possible. I for the function

Compute the difference quotient answer as much as possible for the function f(x) = x + 9.

Simplify your answer as much as possible: (f(x+h) - f(x)) / h

Given that f(x) = x + 9, then f(x+h) = (x+h) + 9 = x + h + 9.

The difference quotient becomes:

(f(x+h) - f(x)) / h = [(x + h + 9) - (x + 9)] / h = [x + h + 9 - x - 9] / h = h / h = 1, provided h ≠ 0.

Paper For Above instruction

The difference quotient is a fundamental concept in calculus that measures the average rate of change of a function over a specific interval. It is mathematically expressed as (f(x+h) - f(x)) / h, where h represents a small change in x. In this case, the function provided is f(x) = x + 9. To compute the difference quotient, we substitute the function into the formula and simplify.

For the function f(x) = x + 9, calculating f(x+h) involves replacing x with x+h, which yields f(x+h) = (x+h) + 9 = x + h + 9. Substituting into the difference quotient formula, we get:

(f(x+h) - f(x)) / h = [(x + h + 9) - (x + 9)] / h. Simplifying the numerator leads to:

x + h + 9 - x - 9 = h. Therefore, the difference quotient simplifies to:

h / h. When h ≠ 0, the h terms cancel out, leaving us with a value of 1. This result indicates that the average rate of change of the function f(x) = x + 9 over any interval is constant and equal to 1. This aligns with the understanding that the function represents a linear relationship with a slope of 1, showing a consistent increase of 1 unit in f(x) for each 1-unit increase in x. The simplicity of this function exemplifies the straightforward nature of the difference quotient for linear functions, where the average rate of change is constant across the domain.

Additional Calculations and Context

If we consider a different function, such as f(x) = -4x^2 - 3x, the difference quotient would involve more complex algebraic manipulation. The process remains the same: substitute f(x+h) and f(x), then simplify. For quadratic functions, the quotient typically yields an expression that varies with x and h, leading to the derivative as h approaches zero.

Similarly, understanding the average rate of change through the difference quotient extends to applications in economics, physics, and other sciences. For example, in economics, the utility function U(x) might represent consumer satisfaction based on the number of units consumed. Calculating the average rate of change over different intervals provides insights into how consumer satisfaction evolves as consumption increases.

In conclusion, computing the difference quotient for various functions helps us understand the concept of derivatives and the behavior of functions. For the specific function f(x) = x + 9, the difference quotient simplifies elegantly to 1, demonstrating the linear nature of the function and its constant rate of change across the domain.

References

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