Question

For \(f(x)=2 x^{2}-1\), find and simplify \([f(a+h)-\) \(f(a)] / h\)

Short answer

\(4a + 2h\)

Step-by-step solution

Evaluate f(a)

First, substitute \(a\) into the function \(f(x) = 2x^2 - 1\). Thus, \(f(a) = 2a^2 - 1\).

Evaluate f(a+h)

Next, substitute \(a + h\) into the function. We have \(f(a + h) = 2(a + h)^2 - 1\). Expand \((a + h)^2\) to get \(a^2 + 2ah + h^2\), so \(f(a + h) = 2(a^2 + 2ah + h^2) - 1\), which simplifies to \(2a^2 + 4ah + 2h^2 - 1\).

Substitute into the Difference Quotient

Substitute \(f(a)\) and \(f(a+h)\) into the expression \([f(a+h) - f(a)] / h\). We get: \[ \frac{2a^2 + 4ah + 2h^2 - 1 - (2a^2 - 1)}{h} \].

Simplify the Expression

Simplify the numerator: \(2a^2 + 4ah + 2h^2 - 1 - 2a^2 + 1 = 4ah + 2h^2\). The expression becomes \(\frac{4ah + 2h^2}{h}\).

Factor and Cancel h

Factor \(h\) out of the numerator: \(h(4a + 2h)\). Cancel \(h\) in the numerator and denominator to simplify: \(4a + 2h\).

Key concepts

Calculus

Calculus is a branch of mathematics focused on understanding how things change. It is all about limits, or in other words, how a function behaves as you approach a certain point. Calculus allows us to study the motion and change and is essential for solving complex problems in science, engineering, and economics.

One of the fundamental tools in calculus is the concept of the difference quotient. This term may sound complex at first, but it essentially represents the average rate of change of a function over a small interval. The difference quotient helps lay the groundwork for derivatives, which provide precise measurements of change at any given point.

• The difference quotient formula is given by: \ \( \frac{f(a+h) - f(a)}{h} \).
• This formula estimates the slope of the tangent line to the curve of a function at a particular point.
• As \( h \) approaches zero, the difference quotient gets closer to the function's derivative.

In our example, we've applied the difference quotient to the function \( f(x) = 2x^2 - 1 \). Understanding this foundational concept is crucial for mastering calculus and unlocking the potential of mathematical analysis in continuous functions.

Derivatives

Derivatives are a core part of calculus, providing us with an exact measure of how a function changes at a specific point. Essentially, if you want to know how fast something is moving at a particular moment, you're thinking about its derivative.

Derivatives extend the concept of the difference quotient to its limit, capturing an instantaneous rate of change. When you take the derivative of a function using the difference quotient approach, you are performing a critical calculation to find this rate.

• The process involves substituting \( a + h \) into the function and then subtracting \( f(a) \), as seen in our worked-out example.
• Dividing this expression by \( h \) and simplifying, gives you insights into how changes in \( x \) affect \( f(x) \).
• For \( f(x) = 2x^2 - 1 \), the simplified derivative is \( 4a + 2h \), representing how the function slope changes as we approach point \( a \).

Mastering derivatives allows you to solve problems related to slopes, tangents, and rates of change, which is invaluable in fields such as physics and economics.

Function Simplification

Function simplification is a critical skill in mathematics, particularly when dealing with complex expressions. Simplifying allows for the clearer understanding and easier handling of functions, which is necessary in deriving more workably refined solutions.

In the given exercise, function simplification involves breaking down complex expressions into manageable parts. This is often achieved through factoring, expanding brackets, and canceling terms.

• First, we substituted \( (a + h) \) into the original function \( f(x) = 2x^2 - 1 \), and expanded it to simplify \( f(a+h) \).
• By subtracting \( f(a) \) and simplifying, complex terms were reduced to \( 4ah + 2h^2 \).
• Factoring out \( h \) from this expression further simplified it to \( h(4a + 2h) \), allowing us to cancel \( h \) from the numerator and the denominator.
• This process left us with the simplified expression \( 4a + 2h \), which is easier to work with, especially when approaching derivatives.

By practicing function simplification, students prepare themselves for advanced calculus tasks and enhance their problem-solving efficiency.