Problem 15 - 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. \(h(x) = \sqrt{x}, \quad (9, 3)\)

Step-by-Step Solution

Answer

The slope of the graph of the function \(h(x) = \sqrt{x}\) at the point (9, 3) is \(1/6\).

Step 1: Understanding the Problem

The goal is to compute the slope of the graph of the function \(h(x) = \sqrt{x}\) at the point (9, 3). We will use the limit definition of the derivative, substituting x = 9.

Step 2: Applying the Limit Definition of the Derivative

We can write the limit definition of the derivative as follows: \(h'(x) = \lim_{h->0} \frac{\sqrt{x + h} - \sqrt{x}}{h}\). Now, we substitute x = 9 in this formula: \(h'(9) = \lim_{h->0} \frac{\sqrt{9 + h} - 3}{h}\).

Step 3: Simplifying the Limit Expression

Before proceeding with the limit computation, it's beneficial to simplify the expression. To do this, we rationalize the numerator: \(h'(9) = \lim_{h->0} \frac{\sqrt{9 + h} - 3}{h} * \frac{\sqrt{9 + h} + 3}{\sqrt{9 + h} + 3} = \lim_{h->0} \frac{9+h-9}{h(\sqrt{9 + h}+3)} = \lim_{h->0} \frac{h}{h(\sqrt{9 + h}+3)}\). After canceling out like terms, we are left with \(h'(9) = \lim_{h->0} \frac{1}{\sqrt{9 + h}+3}\).

Step 4: Evaluating the Limit

The limit can be evaluated by substituting h = 0 in the expression. We get \(h'(9) = \frac{1}{\sqrt{9 + 0}+3} = \frac{1}{3+3} = \frac{1}{6}\).

Key Concepts

Slope of a Function

The slope of a function at a particular point gives us the rate of change at that specific location on the graph. Imagine you're standing on a hill—slope tells us how steep the hill is at any point. In mathematics, this slope enables us to understand how one variable changes with respect to another. For example, if a car is speeding up, the slope of the distance function concerning time tells us how quickly the distance changes. The slope of the function can also signify whether the function is increasing or decreasing at that point. If the slope is positive, the function is rising; if negative, it is falling. To find the slope of a function at a given point, we use the derivative from calculus, which provides an analytical way to determine the slope with precision.

Derivative

The derivative of a function is a powerful concept in calculus that helps us find the slope of the tangent line at any point on a graph. Think of the derivative as a tool that gives us the exact rate of change of the function at that point. Finding the derivative involves a limit process. For the function, say, \(h(x) = \sqrt{x}\), its derivative at a specific point \(x\) is given by the formula: \[h'(x) = \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}\]This expression helps capture how the function values change at infinitesimally small distances. A derivative is important because it tells us much more than just the slope. It can convey whether the graph is flattening out, peaking, or dipping. Additionally, derivatives are foundational in many fields, from physics (to study motion) to economics (to understand cost functions).

Rationalizing

Rationalizing is a technique we use to simplify expressions, especially when working with radicals like square roots in the numerator. This method helps in solving complex limit problems by making them more manageable. When taking derivatives, it often appears in the limit definition when simplifying expressions. For instance, in our exercise with \(h(x) = \sqrt{x}\) at \(x = 9\), we rationalize to make calculations smoother. The tricky expression \(\frac{\sqrt{9 + h} - 3}{h}\) becomes easier to handle by multiplying both the numerator and the denominator by \(\sqrt{9 + h} + 3\), leading us to:\[\frac{(\sqrt{9 + h} - 3)(\sqrt{9 + h} + 3)}{h(\sqrt{9 + h} + 3)}\] This simplifies the squaring of terms and allows us to eventually cancel out \(h\) in the fraction, making the limit evaluation possible and approachable.

Evaluating Limits

Evaluating limits is a fundamental process in calculus that allows us to find the value a function approaches as the input gets arbitrarily close to a specific point. Think of limits as understanding what happens at the "boundary" or "extremes" of a function. In our exercise, after simplifying the expression with rationalizing, we end up with a manageable limit: \[\lim_{h \to 0} \frac{1}{\sqrt{9 + h}+3}\]To evaluate this, we substituted \(h = 0\) into the expression, making it easy to see the function's behavior without direct computation of undefined situations like division by zero. Evaluating limits is crucial for determining the derivative, which, as we've noted, signifies the slope of a function at a point. This concept underpins many mathematical theories and applications, providing insight into behavior not just at specific points, but also in trends and models across various scientific and engineering disciplines.