Let g(x)=√(x) for x ≥0 a. Find the average rate of change of...

Let $g(x)=\sqrt{x}$ for $x \geq 0$ a. Find the average rate of change of $g(x)$ with respect to $x$ owerthe 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 $[1,1+h]$ .

Let $g(x)=\sqrt{x}$ for $x \geq 0$
a. Find the average rate of change of $g(x)$ with respect to $x$ owerthe 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 $[1,1+h]$ .

Key Concepts

Average Rate of Change

The average rate of change quantifies how a function's value changes over a specified interval, calculated by (f(b) - f(a))/(b - a). This concept underlies the idea of a secant line's slope and serves as a precursor to understanding instantaneous changes.

Difference Quotient

The difference quotient, expressed as (f(x + h) - f(x))/h, measures the average rate of change over an interval from x to x+h. It is the foundational expression used when transitioning towards the definition of the derivative through a limiting process.

Limit Process

The limit process involves determining the value that a function approaches as the input nears a particular point. In calculus, it is essential for evaluating the instantaneous rate of change of a function, as seen when the interval size h tends to zero.

Instantaneous Rate of Change (Derivative)

The derivative represents the instantaneous rate of change, indicating how fast a function changes at a specific point. It is defined as the limit of the difference quotient as h approaches zero, thus linking average change over an interval to the exact change at a point.

Numerical Approximation and Tabular Analysis

Creating a table of average rates of change for progressively smaller intervals is a numerical method to approximate the derivative. This approach demonstrates how, as the interval decreases, the average rate of change converges to the derivative, offering a practical insight into the behavior of functions as they approach instantaneous change.

Transcript

00:01 All right here we're going to do problem 19 from chapter 2, limits and continuity, section 1, rates of change and tangents to curves from the textbook thomas calculus.

00:13 We have a function here.

00:15 We want to let g of x equal the square root of x for all x greater than or equal to zero.

00:23 So we're just dealing with real numbers.

00:26 And we want to find the average rate of change with g of x, with right of change with right.

00:31 Respect over these intervals from 1 to 2, from 1 to 1 .5, and from 1 to 1 plus h.

00:49 Okay, so we have these intervals.

00:51 So remember we're going for our x values 1 to 2.

00:54 So that means i need to know g of 1 is equal to what, the square root of 1, which is equal to 1, g of 2 is equal to the square root of 2, which is just the square root 2.

01:13 So then to find the average rate of change here, i just need to find the slope.

01:19 The slope is going to be square root of 2 minus 1 divided by 2 minus 1.

01:28 So we're going to have to kind of estimate here because square root of 2, of course, is an irrational number.

01:34 I grab my calculator.

01:37 You probably heard that click.

01:38 Just grabbing my calculator here.

01:41 So i've got square root of two minus one and that is over two minus one.

01:53 And that gives me about 0 .414 .2.

01:59 That's rounded, of course.

02:03 All right.

02:03 And then we've got our second interval.

02:04 Again, it's g of 1.

02:06 We already know that's equal to 1.

02:08 And then g of 1 .5, we just is going to be equal to the square root of 1 .5.

02:14 I'm going to leave it right there for now, and i'm going to wait until i calculate this slope to round.

02:21 So that's going to give me the square root of 1 .5 minus 1 over 1 .5 minus 1.

02:32 And when i calculate that out, i get 0 .4495.

02:41 And again, that's rounded, of course.

02:42 All right, i've got one more.

02:45 Now here, since i've got 1 and 1 plus h, i'm going to use that difference quotient that we talked about before to get the slope there, which is f of x plus h minus f of x over h.

03:03 So in this case, x is 1.

03:05 So we're going to do f of 1 plus h minus f of 1 divided by 8.

03:16 So let me give myself a little bit of room here.

03:21 So this is going to give me the square root of 1 plus h minus the square root of 1, which is 1 over h.

03:36 So that is that slope since we don't know what h is.

03:39 But the second part of this question asks us to make a table of values with respect to this interval and you plug in some some values for h.

03:48 We can't plug in zero because we have h in a denominator.

03:52 We can't have zero in the denominator.

03:55 So i would say h and then here we're going to, let's say we want 0 .1.

04:05 I'm just adding a zero each time up to a couple more.

04:22 Now below here i'm going to put h plus 1, which is going to be pretty simple to calculate.

04:29 We're just going to add...