Problem 39 - Algebra and Trigonometry

Question

$$\text { If } f(x)=5 x^{2}-6 x+1, \text { find } \frac{f(x+h)-f(x)}{h}$$ (Section 2.2, \text { Example } 8)

Step-by-Step Solution

Answer

After calculating and simplifying the difference quotient, the final answer is \(10x-6\).

Step 1: Substitute \(x+h\) into the function

First we calculate \(f(x+h)\). For this, we replace every \(x\) in the function with \(x+h\): \(f(x+h) = 5(x+h)^2 - 6(x+h) + 1\).

Step 2: Simplify \(f(x+h)\)

Expand and simplify the expression: \(f(x+h) = 5(x^2+2hx+h^2) -6x -6h +1 = 5x^2+10hx+5h^2-6x-6h+1.\)

Step 3: Compute the difference \(f(x+h)-f(x)\)

Now we need to subtract \(f(x)\) from \(f(x+h)\). This gives: \(f(x+h) - f(x) = (5x^2+10hx+5h^2-6x-6h+1) -(5x^2-6x+1) = 10hx+5h^2-6h.\)

Step 4: Divide by \(h\)

Now we divide the difference \(f(x+h)-f(x)\) by \(h\): \(\frac{f(x+h)-f(x)}{h} = \frac{10hx+5h^2-6h }{h} =10x+5h-6.\)

Step 5: Final simplification

Finally, we simplify the expression. Since \(h\) is just a small increment to \(x\) and will approach zero, our final answer becomes \(10x-6\).

Key Concepts

Average Rate of Change

The average rate of change of a function is a concept that measures how much the function's output changes with respect to a change in the input. Think of it as the function's average speed over a certain interval. For the function given by the exercise, f(x) = 5x^2 - 6x + 1, the average rate of change when the input changes from x to x+h is given by the difference quotient, \(\frac{f(x+h)-f(x)}{h}\).

This quotient represents the slope of the secant line that intersects the graph of the function at points (x, f(x)) and (x+h, f(x+h)). The closer h gets to zero, the more the average rate of change resembles the instantaneous rate of change at the point x, which is the derivative of the function at x.

Polynomial Functions

Polynomial functions are mathematical expressions that involve sums of powers of the variable. The function f(x) = 5x^2 - 6x + 1 is an example of a second-degree polynomial, commonly known as a quadratic function. Quadratics are characterized by the highest exponent being 2. They generally have a parabolic shape when plotted on a graph.

Polynomial functions are useful due to their simplicity and the ease with which we can perform operations such as addition, subtraction, and finding derivatives. For example, in the given exercise, the polynomial made it straightforward to compute f(x+h) and then to find the difference quotient.

Function Transformation

Function transformation involves shifting, stretching, compressing, or reflecting a function's graph. When we substitute x with x+h in our function, we are essentially shifting the graph of f(x) horizontally. This does not change the shape of the graph but moves it along the x-axis.

If h > 0, the graph shifts to the left, and if h < 0, it shifts to the right. The value of h in the difference quotient does not transform the function permanently; it is just part of a process to understand the function's behavior at and around the point x.

Algebraic Simplification

Algebraic simplification is the process of making an expression simpler and more manageable without changing its value. It often involves combining like terms, factoring, expanding, or canceling common factors. In the exercise, after finding f(x+h), we simplified the expression by expanding it and combining like terms.

In the step-by-step solution provided, after computing the difference f(x+h) - f(x) and dividing by h, we simplified the result to 10x + 5h - 6. Then, acknowledging that h approaches zero, the term 5h is dropped which gives us the simplified final expression 10x - 6. This last step is crucial in preparing the expression for further calculus operations, such as taking the limit as h approaches zero.