Problem 12 - Precalculus with Limits

Question

In Exercises 9-16, use the limit process to find the slope of the graph of the function at the specified point. Use a graphing utility to confirm your result. \(h(x) = 2x+5, \quad (-1, 3)\)

Step-by-Step Solution

Answer

The slope of the function \(h(x) = 2x+5\) at the point \(-1, 3\) is \(2\).

Step 1: Substitute Function in the Limit Definition of a Derivative

We first substitute our function and the point into the limit definition of a derivative formula. \[f'(a)= \lim_{{h \to 0}} \frac{{f(a+h)-f(a)}}{{h}}\] This gives us: \[h'(-1)= \lim_{{h \to 0}} \frac{{h(-1+h)-h(-1)}}{{h}}\] We simplify this to: \[h'(-1)= \lim_{{h \to 0}} \frac{{h(-1+h)-h(-1)}}{{h}}\]

Step 2: Simplify the Expression

The next step is to simplify the above expression. The result is \[h'(-1)= \lim_{{h \to 0}} \frac{{2h-2}}{h}\]

Step 3: Reduce the Function

From here, reduce the function \[h'(-1)= 2 \]

Step 4: Verify using a Graphing Utility

This would be confirmed by utilising a graphing tool to plot the function \(h(x) = 2x +5\), and inspecting the slope at the point \(-1, 3\). The slope of the tangent line at that point should be \(2\).

Key Concepts

Limit Process

The limit process is a fundamental concept in calculus used to find the derivative of a function at a particular point. The derivative represents the slope of the tangent line to the graph at that specific point. In our case, we are given the function \(h(x) = 2x + 5\) and need to find its slope at the point \((-1, 3)\). We do this using the limit definition of a derivative, which is given by:
\[f'(a) = \lim_{{h \to 0}} \frac{{f(a+h)-f(a)}}{h}\]
This formula essentially looks at the average rate of change of the function as the interval \(h\) approaches zero, thus capturing the instantaneous rate of change, or the slope, at the specific point. It is crucial to substitute the function and points correctly to mirror the behavior of the function near the desired point.

Slope of a Graph

The slope of a graph is a measure of how steep the line is at any given point. For a linear function like \(h(x) = 2x + 5\), the slope is constant along the entire graph. However, when dealing with nonlinear functions, the slope can vary. That's where finding the derivative using the limit process becomes invaluable. The slope at a point on the graph is calculated by taking the derivative at that point. In our exercise, after applying the limit process, we found that the derivative \(h'(-1)\) equals 2. This value indicates that at the point \((-1, 3)\), the graph rises 2 units vertically for every unit it moves horizontally. A positive slope like this suggests an upward incline on the graph.

Tangent Line

A tangent line to a graph at a certain point is a straight line that just "touches" the graph at that point and has the same slope as the graph does at that point. It provides a good linear approximation of the graph close to the point of tangency. For our problem, the tangent line to \(h(x) = 2x + 5\) at the point \((-1, 3)\) would have a slope of 2, as derived through the limit process. Understanding tangent lines is essential because they help in visualizing how the function behaves at a given point. They are especially useful when approximating functions or exploring instantaneous rates of change in real-world scenarios.

Graphing Utility

A graphing utility is a helpful tool to confirm mathematical calculations visually. It allows us to plot functions and inspect their behavior graphically. In this exercise, we use a graphing utility to ensure the slope calculated by the limit process matches the visual slope of the tangent line on the graph of \(h(x) = 2x + 5\). By inputting the function into the graphing tool and locating the point \((-1, 3)\), the utility should display a line with a slope of 2, confirming our previous analytical calculations. Utilizing technology in this way provides a visual reinforcement of the mathematical concepts, ensuring a deeper understanding of the interplay between algebraic and graphical interpretations.