Question

Simplify the difference quotients\(\frac{f(x+h)-f(x)}{h}\) and \(\frac{f(x)-f(a)}{x-a}\) for the following functions. $$f(x)=2 x^{2}-3 x+1$$

Short answer

a) \(\frac{f(x+h)-f(x)}{h}\) b) \(\frac{f(x)-f(a)}{x-a}\) Answer: a) \(4x+2h-3\) b) \(2(x+a)-3\)

Step-by-step solution

Calculate \(f(x+h)\)

Replace \(x\) with \(x+h\) in the given function: $$f(x+h)=2 (x+h)^{2}-3 (x+h)+1$$

Expand and simplify \(f(x+h)\)

Expand and combine like terms: $$f(x+h)=2(x^2+2xh+h^2)-3x-3h+1$$ $$f(x+h)=2x^2+4xh+2h^2-3x-3h+1$$

Calculate \(f(x+h)-f(x)\)

Subtract \(f(x)\) from \(f(x+h)\) to find the difference: $$(f(x+h)-f(x))=(2x^2+4xh+2h^2-3x-3h+1)-(2x^2-3x+1)$$ Simplify by removing the brackets and then subtracting the corresponding terms: $$f(x+h)-f(x)=4xh+2h^2-3h$$

Simplify the first difference quotient \(\frac{f(x+h)-f(x)}{h}\)

Divide the difference \(f(x+h)-f(x)\) by \(h\): $$\frac{f(x+h)-f(x)}{h}=\frac{4xh+2h^2-3h}{h}$$ Factor out an \(h\) from the numerator and simplify: $$\frac{f(x+h)-f(x)}{h}=\frac{h(4x+2h-3)}{h}=4x+2h-3$$ Now we'll work with the second difference quotient:

Calculate \(f(x)-f(a)\)

Replace \(x\) with \(a\) in the given function and subtract it from \(f(x)\): $$(f(x)-f(a))=(2x^2-3x+1)-(2a^2-3a+1)$$ Simplify by removing the brackets and subtracting corresponding terms: $$f(x)-f(a)=2x^2-2a^2-3x+3a$$

Simplify the second difference quotient \(\frac{f(x)-f(a)}{x-a}\)

Divide the difference \(f(x)-f(a)\) by \(x-a\): $$\frac{f(x)-f(a)}{x-a}=\frac{2x^2-2a^2-3x+3a}{x-a}$$ Factor out a 2 from the first two terms and a 3 from the last two terms in the numerator: $$\frac{f(x)-f(a)}{x-a}=\frac{2(x^2-a^2)-3(x-a)}{x-a}$$ Now, notice that the first term in the numerator is a difference of squares, and the second term is a common factor. Simplify further: $$\frac{f(x)-f(a)}{x-a}=\frac{2(x-a)(x+a)-3(x-a)}{x-a}$$ Factor \((x-a)\) from the numerator: $$\frac{f(x)-f(a)}{x-a}=2(x+a)-3$$ Now we have simplified both difference quotients: $$\frac{f(x+h)-f(x)}{h}=4x+2h-3$$ $$\frac{f(x)-f(a)}{x-a}=2(x+a)-3$$

Key concepts

Derivative

In calculus, the concept of a derivative is fundamental. It measures the rate at which a function changes at any given point. Essentially, derivatives give us the slope of the tangent line to a function's graph at a specific point.
The formal definition of a derivative at a point is given by the limit of the difference quotient:

• As the change in input (often represented as \(h\) or \(x-a\)) approaches zero.
• This results in an instantaneous rate of change.

If we have a function \(f(x)\), its derivative, often written as \(f'(x)\), is calculated as: \[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h)-f(x)}}{h}\]For the exercise, simplifying the difference quotients helps to conceptually approximate derivatives, as it captures how the function behaves as inputs shift ever so slightly.

Function Simplification

To grasp the behavior of functions, simplifying complex expressions is often necessary. Function simplification involves reducing expressions to a more manageable form without altering their intrinsic value.
This is accomplished by:

• Combining like terms
• Expanding expressions
• Cancelling common factors

In the given exercise, we start with substituting \(x+h\) into the function to find \(f(x+h)\). This is important in approaching the difference quotient calculation. By expanding \(f(x+h)\) and simplifying the resultant expression, calculation becomes more straightforward. This key operation not only helps in derivatives but also in other aspects such as integration and solving equations.

Algebraic Manipulation

Algebraic manipulation is a powerful tool in mathematics that involves rearranging and simplifying expressions using algebraic rules. This skill is crucial in calculus, especially when dealing with complex functions and expressions.

• Basic operations include expanding expressions, factoring, and simplifying fractions.
• For expressions containing a difference of squares, recognizing this form aids in swift simplification.

In the problem at hand, after forming the difference formula, algebraic manipulation becomes essential. Notice in the step where \(f(x)-f(a)\) was simplified by utilizing the difference of squares formula:\[ 2(x^2 - a^2) = 2(x-a)(x+a) \]These manipulations allow us to factor out terms effectively, leading to a neat and simplified expression. Mastery of these techniques is necessary for progressing in calculus or any higher-level mathematics tasks.