Problem 19 - Thomas Calculus

Question

Let \(g(x)=\sqrt{x}\) for \(x \geq 0.\) a. Find the average rate of change of \(g(x)\) with respect to \(x\) over the intervals \([1,2],[1,1.5]\) and \([1,1+h] .\) b. Make a table of values of the average rate of change of \(g\) with respect to \(x\) over the interval \([1,1+h]\) for some values of \(h\) approaching zero, say \(h=0.1,0.01,0.001,0.0001,0.00001,\) and \(0.000001 .\) c. What does your table indicate is the rate of change of \(g(x)\) with respect to \(x\) at \(x=1 ?\) d. Calculate the limit as \(h\) approaches zero of the average rate of change of \(g(x)\) with respect to \(x\) over the interval \([1,1+h].\)

Step-by-Step Solution

Answer

The rate of change at \( x = 1 \) for \( g(x) \) is \( \frac{1}{2} \).

Step 1: Find Average Rate of Change for [1, 2]

The average rate of change of a function over an interval \([a, b]\) is given by \((f(b) - f(a)) / (b - a)\). For the function \([g(x)=\sqrt{x}]\):\(g(1) = \sqrt{1} = 1\), \([g(2) = \sqrt{2}]\).Calculating the rate: \[\frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1.\\]

Step 2: Find Average Rate of Change for [1, 1.5]

Similarly, calculate for \([1, 1.5]\): \([g(1.5) = \sqrt{1.5}]\).The rate is: \[\frac{\sqrt{1.5} - 1}{1.5 - 1} = \frac{\sqrt{1.5} - 1}{0.5}.\\]

Step 3: Find Average Rate of Change for [1, 1 + h]

For the interval \([1, 1+h]\), the average rate of change is given by: \[\frac{\sqrt{1+h} - 1}{h}.\]

Step 4: Create the Table for Various Values of h

Calculate the average rate for small values of \(h\): 0.1, 0.01, 0.001, 0.0001, 0.00001, 0.000001: \[\frac{\sqrt{1.1} - 1}{0.1}, \frac{\sqrt{1.01} - 1}{0.01}, \frac{\sqrt{1.001} - 1}{0.001}, \frac{\sqrt{1.0001} - 1}{0.0001}, \frac{\sqrt{1.00001} - 1}{0.00001}, \frac{\sqrt{1.000001} - 1}{0.000001}.\]

Step 5: Rate of Change at x = 1

Observe from the table that as \(h\) approaches 0, the average rate values seem to approach a specific number:\[\text{Limit } \underset{h \to 0}{\lim} \frac{\sqrt{1+h} - 1}{h} \approx 0.5.\]

Step 6: Calculate the Limit

To find \(\lim_{h \to 0} \frac{\sqrt{1+h} - 1}{h}\), apply the binomial expansion or rationalization technique:Rationalizing the numerator, multiply by the conjugate:\[\frac{\sqrt{1+h} - 1}{h} \times \frac{\sqrt{1+h} + 1}{\sqrt{1+h} + 1} = \frac{h}{h (\sqrt{1+h} + 1)}.\]Then:\[\lim_{h \to 0} \frac{1}{\sqrt{1+h} + 1} = \frac{1}{2}.\]

Key Concepts

Derivatives

In calculus, derivatives are quite essential as they express the rate at which a function changes. When you think of derivatives, think of them as the tool that tells us how steep a function is at any point. This is particularly important when we're interested in the instantaneous rate of change, which contrasts with the average rate of change over an interval.

To connect this with the given problem, the average rate of change of the function \(g(x)=\sqrt{x}\) over an interval \([a, b]\) is calculated using:

• \(f(b) - f(a)\)
• divided by \(b - a\)

This average rate helps approximate how fast the function is changing over an interval. But, when we're looping in derivatives, we're zooming in more, focusing on what happens as the interval becomes infinitely small. This is exactly what the derivative at a point gives us – the instantaneous rate of change at that very point.

In essence, in problems like the one with \(g(x)=\sqrt{x}\), the process of narrowing down the average rate of change (e.g., by decreasing \(h\) in \([1, 1+h]\)) bridges directly to the derivative's core concept.

Limits

Limits are the fundamental building blocks to understanding so much in calculus, including derivatives. They allow us to define what happens to the behavior of a function as we approach a specific point or as the input becomes very large or very small.

For parts of this exercise, we're dealing with the limit as \(h\) approaches zero in the expression \(\frac{\sqrt{1+h} - 1}{h}\). This concept is all about seeing what's going on as \(h\) gets closer and closer to zero, without actually being zero – a mirror to real-world problems where we are observing tendencies rather than immediate values.

The limit we calculate here helps us pinpoint the exact rate of change of \(g(x)=\sqrt{x}\) at \(x=1\). The rationalization step, where we multiply by the conjugate, is one common technique to evaluate limits involving roots. As \(h\) goes to zero, we see the analytical value settling down to \(0.5\), which signifies the slope of \(g(x)\) at that point.

Continuous Functions

Understanding continuous functions is crucial in calculus because their traits allow us to talk about derivatives and limits sensibly. A continuous function is one that doesn't "jump" or break apart; its graph can be drawn without lifting your pen from the paper.

In the exercise with \(g(x)=\sqrt{x}\), we're dealing with a function that is continuous for \(x \geq 0\). This continuity permits us to investigate the rate of change, as the transitions between any two points are seamless. When functions are continuous, limits and derivatives make sense since there are no unexpected gaps or discrepancies in values.

Continuous functions like \(g(x)=\sqrt{x}\) give us the perfect playground for exploring rates of change smoothly, and that's one reason this kind of function is so often used in introductory calculus exercises. Identifying this continuity underpins why we can readily utilize limits to find derivatives across intervals.