Problem 72 - Calculus Volume 1

Question

For the following exercises, the given limit represents the derivative of a function \(y=f(x)\) at \(x=a .\) Find \(f(x)\) and \(a .\) $$\lim _{h \rightarrow 0} \frac{\left[2(3+h)^{2}-(3+h)\right]-15}{h}$$

Step-by-Step Solution

Answer

The function is \( f(x) = 2x^2 - x \) and \( a = 3 \).

Step 1: Identify the derivative form

The given limit is in the form of a derivative definition: \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \). This is used to find the derivative of the function \( f(x) \) at \( x = a \). The expression in the limit closely matches this form.

Step 2: Match the form to identify \( a \)

Look at the expression \( [2(3+h)^2 - (3+h)] - 15 \). From this, recognize that \( 3 + h \) is used in the same way as \( a + h \) in \( f(a+h) \). Thus, we find that \( a = 3 \).

Step 3: Identify \( f(x) \)

Rewrite \( f(a+h) = 2(3+h)^2 - (3+h) \). Replace \( 3 + h \) with \( x \) to find \( f(x) = 2x^2 - x \). Thus, \( f(x) = 2x^2 - x \).

Step 4: Verify the form and solution

Substituting back, we have \( f(3) = 2(3)^2 - 3 = 18 - 3 = 15 \). When \( a = 3 \), substitute \( f(x) \) back into the definition: the limit \( \lim_{h \to 0} \frac{f(3+h) - 15}{h} \) matches \( \lim_{h \to 0} \frac{2(3+h)^2 - (3+h) - 15}{h} \). This confirms the function \( f(x) = 2x^2 - x \) is correct.

Key Concepts

Derivative

A derivative is a fundamental concept in calculus. It is essentially about finding the rate at which a function changes at any given point. Formally, the derivative of a function \( f(x) \) at a point \( x = a \) is given by the limit:

• \( \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \)

This formula allows us to understand how the function behaves as \( x \) approaches \( a \).
In our problem, recognizing this derivative form helped us identify \( a = 3 \) and \( f(x) = 2x^2 - x \). This expression within the limit matched our known form, thus identifying the respective function.

Limits

Limits in calculus help us understand the behavior of functions as they approach specific points or infinity. The limit is foundational for defining derivatives. In our exercise, we used a limit to determine the derivative of \( f(x) \).
The limit format \( \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \) becomes crucial. We compute limits to evaluate how a function responds when a variable increment (here \( h \)) becomes very small.
Understanding limits allows you to probe into the function's behavior around a critical value. So, whether we are interested in finding rates of change or understanding continuity, limits are essential.

Function Behavior

Analyzing a function's behavior involves understanding how it changes over its domain. The behavior of a function can be indicated by where it increases, decreases, or remains constant.
With derivatives, we express these changes quantitatively. The derivative \( f'(x) \) at a point tells us the slope of the tangent line to the graph at that point.

• A positive derivative means the function is increasing.
• A negative derivative means it is decreasing.

In our case, \( f(x) = 2x^2 - x \) describes a quadratic function. Such functions exhibit parabolic behavior, which means analyzing its derivative can tell us about its turning points and concavity.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying expressions to provoke understanding and solve mathematical problems. This is crucial in calculating derivatives and evaluating limits.
In the exercise, identifying \( f(a+h) = 2(3+h)^2 - (3+h) \) required expanding \((3+h)^2\) and combining terms.

• Expanding: \( (3+h)^2 = 9 + 6h + h^2 \)
• Applying the coefficients: \( 2(9 + 6h + h^2) \)

Such manipulations clarify and simplify the problem, making the structure visible. Without these skills, extracting function information from complex expressions would be much harder.