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

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 $[1,1+h].$

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 $[1,1+h].$

Key Concepts

Derivative

The derivative of a function at a point is defined as the limit of the difference quotient as the change in input approaches zero. It provides a precise measure of the function’s instantaneous rate of change, indicating the slope of the tangent line to the function's graph at that point. This concept is essential in calculus, linking the ideas of limits, continuity, and rates of change.

Limit of a Function

A limit is a fundamental concept in calculus that describes the behavior of a function as its input approaches a particular value. When computing the instantaneous rate of change (or derivative), limits are used to determine what the average rate of change approaches as the interval becomes infinitesimally small. This concept is key in transitioning from an average rate over an interval to the exact rate at a specific point.

Instantaneous Rate of Change

The instantaneous rate of change of a function at a given input is the value that the average rate of change approaches as the interval around that input shrinks to zero. It represents how quickly the function's output is changing at that exact moment and is computed via the limit of the difference quotient, which is the essence of the derivative.

Average Rate of Change

The average rate of change of a function over an interval is a measure of how much the function's value changes per unit change in the input over that interval. It is computed by taking the difference in the function values at the endpoints of the interval and dividing by the difference in the input values. This concept lays the groundwork for understanding how functions behave over larger intervals.

Difference Quotient

The difference quotient is the expression formed by dividing the change in the function’s value by the change in the input value over a small interval. In mathematical terms, it is often written as [f(x+h) – f(x)]/h. This quotient is central to computing the average rate of change and serves as the foundation for defining the derivative as h approaches zero.

Transcript

00:05 All right, so the problem we're going to be working on is going to be giving us a function.

00:11 G of x is equal to the square of x.

00:14 And it's going to ask us a couple of questions.

00:17 So question a asks us for the rate of change, the average rate of change, of g of x with respect to x over these three intervals that i've listed here.

00:26 So the basic equation for average rate of change is the value of the function at one point.

00:30 So that's g of xf minus the value of the function at another point.

00:34 I'm calling this g of xi over the change in that interval.

00:39 So we have xf minus xi.

00:41 So let's just go through this.

00:44 So our xf is going to be the second point, and our xi is going to be the first point.

00:47 So if we do it for this first interval here, 1 -2, we're going to have square root of 2 minus square root of 1, all over 2 minus 1, right? and this is just, if we simplify, it's just square root of 2.

01:06 To minus 1.

01:10 Okay.

01:10 So that is the answer for the first question for the first interval.

01:14 The second one, if we do the same thing, let me just write, just looking at everything, fractions, 1 .5 is equal to three halves, right? so i'm going to write the second part, and it's square root of three halves minus the square root of one.

01:31 I'm sure right.

01:32 It is one.

01:32 We know what that is.

01:34 And then we have 1 .5 minus 1.

01:38 Okay.

01:38 It and then that one point i'm sorry i had minus one so if we rewrite this what we actually can get if we simplify it is the square root of six square to six minus two and you can feel free to leave it leave it in this first form right here you can feel free to leave it there you don't have to if you want to simplify it you can go in this more simplified form and that you just get that from taking square to three over two divided by 0 .5, the denominator, and then minus 1 divided by 0 .5.

02:18 And that minus 1 divided by 0 .5 is just the 2.

02:21 And the square of 3 over 2 divided by 2 is just the square to 6.

02:27 Okay.

02:27 And then for the last one, we have square root.

02:31 That final term is just 1 plus h.

02:34 Subtracted by again, square root 1 is just 1.

02:39 And in the denominator, we have 1 plus h minus 1.

02:44 So we can rewrite this as squared of 1 plus h minus square root of 1 plus h minus 1, all over h.

02:56 That's the answer for our third interval there.

03:00 Okay.

03:01 So now, moving down here, part b asks us to make a table of of the average rate of change of g with respect to x over the interval 1, 1 plus h.

03:13 Well, that 1 ,1 plus h interval, that's what we just defined on right here.