Problem 1 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Which of the following are equal to \(g^{\prime}(3) ?\) A sketch with labeled points will be useful. (a) \(\lim _{h \rightarrow 0} \frac{g(3+h)-g(3)}{h}\) (b) \(\lim _{x \rightarrow 0} \frac{g(x)-g(3)}{x-3}\) (c) \(\lim _{x \rightarrow 3} \frac{g(x)-g(3)}{x-3}\) (d) \(\lim _{s \rightarrow 3} \frac{g(3)-g(s)}{3-s}\) (e) \(\lim _{\Delta x \rightarrow 3} \frac{g(3+\Delta x)-g(3)}{\Delta x}\) (f) \(\lim _{\Delta x \rightarrow 0} \frac{g(3+\Delta x)-g(3)}{\Delta x}\) (g) \(\lim _{\Delta x \rightarrow 0} \frac{g(3)-g(3+\Delta x)}{-\Delta x}\)

Step-by-Step Solution

Answer

The expressions equivalent to \(g'(3)\) are (a), (c) and (f).

Step 1: Identify the derivative formula

The limit definition of the derivative of function g at the point 3 is represented by the following equation: \(g'(3)=\lim_{h \to 0} \frac{g(3+h)-g(3)}{h}\). Therefore any expression equivalent to this will be an answer.

Step 2: Evaluate options

Comparing the given limit expressions with the derivative formula: (a) is the exact formula so this will definitely be equal to \(g'(3)\). (b) is not equivalent because \(x \rightarrow 0\) rather than \(h \rightarrow 0\). (c) is equivalent but the \(x\) is approaching 3 rather than the \(h\) approaching 0, still it's valid as \(x-3 = h\). (d) is the opposite of the formula with the \((3-s) = -h\). The rest expression becomes \(-g'(3)\) so it's not equivalent. (e) not equivalent because it uses \(Δx \rightarrow 3\) instead of \(Δx \rightarrow 0\). (f) is also exactly equivalent to the formula. (g) is same as (f) but with a negative sign. The rest expression becomes \(-g'(3)\) so it's not equivalent.

Key Concepts

Limit Definition of Derivative

Understanding the limit definition of the derivative is crucial in grasping calculus concepts. Put simply, the derivative measures how much a function's output changes as its input changes. The formal definition uses limits to describe the exact rate at which this change happens at a particular point.

Let's take a function, say, g(x). To find the derivative of this function at a point, say x = 3, we need to calculate the limit as 'h' approaches 0 of the expression \(\frac{g(3+h)-g(3)}{h}\). What this does is compare the function's values at very close points, 3 and 3+h, and divide by how close these points are, 'h', as h gets infinitesimally small. Conceptually, it's like zooming in on the function's graph at x=3 so closely that it begins to look straight, and the slope of that line is the derivative.

When given various limit expressions, identifying which ones align with this concept helps determine the correct derivative expression. The exercises (a), (c), (f), and (g) are different variations of the limit definition, but only some maintain the derivative's true meaning due to the approach of the limit.

Calculus

Calculus is a vast field in mathematics that deals with change and motion. It's divided primarily into two branches: differential calculus, concerning the concept of a derivative, and integral calculus, focused on the concept of an integral.

The derivative is one of the fundamental tools in calculus and is used to calculate the rate at which things change. It allows us to determine the velocity of a car if we know its changing position over time, for example. In the exercise, identifying the correct expression for the derivative using the limit definition is an application of differential calculus.

Moreover, calculus is not just about computation; it's a way of thinking. It allows us to model real-world situations, solve practical problems in engineering and physics, and delve into theoretical realms in higher mathematics. Every time we face an equation involving the limit definition of a derivative, we are applying the core principles of calculus.

Function Rate of Change

The rate of change of a function provides us insight into how one quantity changes in relation to another. When we say 'derivative,' we are referring to this very concept of how quickly the function's output (the dependent variable) is changing at a particular input value (the independent variable).

In our exercise, when we try to find \(g'(3)\), we are essentially looking for the rate at which the function \(g(x)\) is changing at the exact point where \(x = 3\). Hence, the solutions (a), (c), and (f) correctly represent this instantaneous rate of change, since they describe the slope of the function at that exact point.

Understanding the rate of change of a function is not just a mathematical exercise. It has practical implications, from calculating the speed of an object, the growth rate of populations, to the sensitivity of investments in finance. It's the concept that bridges the abstract mathematics of calculus with the real-world phenomena that change and evolve over time.