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 $x$ 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 $x=1 ?$

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 $x=1 ?$

Key Concepts

Average Rate of Change

This concept quantifies how much a function's output changes on average per unit change in the input over a specific interval. It is computed by subtracting the function value at the start of the interval from the function value at the end, and then dividing by the difference in the input values.

Difference Quotient

The difference quotient is a formula expressed as (f(x+h) - f(x)) / h. It represents the average rate of change over the interval from x to x+h and serves as the foundational expression for computing derivatives.

Instantaneous Rate of Change

This is the rate at which a function changes at an exact point. It is found by taking the limit of the average rate of change as the interval's length approaches zero, leading to the concept of the derivative.

Limit

A limit describes the behavior of a function as its argument approaches a particular value. In the context of rates of change, limits enable the transition from average to instantaneous measures, forming the core idea behind derivatives.

Transcript

00:01 We're going to find the average rate of change for g of x on a few different intervals and then actually use this to help us maybe decide what might be happening as a rate of change at a moment in time.

00:12 So first of all, let's consider a few different time intervals to check the average rate of change.

00:18 So we're going to look from 1 to 2, from 1 to 1 .5, and then to some kind of unknown endpoint at 1 to 1 to 1 plus.

00:30 So somewhere near 1.

00:34 For the average rate of change, remember that we need to use g of b b minus g of a over b minus a.

00:41 It's like your slope formula.

00:48 So for the first time interval, if i'm looking from 1 to 2, and i'm doing g of 2 minus g of 1 over 2 minus 1, then i'm going to have the square root of two minus the square root of one over one.

01:07 And the square root of two minus one is the most simplified version of that answer to keep it exact.

01:14 So i would say my average rate of change from one to two, the square root of two minus one.

01:19 But to kind of know what's happening here, i'm also going to approximate it.

01:22 And if i approximate this rate of change, i get 0 .41.

01:28 So this is going to help me see maybe what my values are approaching later on this problem.

01:34 On the second interval, i'm going from 1 to 1 .5.

01:37 So i would do g of 1 .5 minus g of 1.

01:42 Over 1 .5 minus 1.

01:49 So then g of 1 .5 is the square root of 1 .5.

01:53 And then g of 1 is a square root of 1.

01:56 Over 1 .5 minus 1 is just 0 .5.

01:59 So again, this is a pretty good simplified version of my answer.

02:04 But to approximate it, it's going to help me know what's happening later.

02:06 And the approximation of this is 0 .449, if i rounded three decimal places.

02:14 So then on my last interval, i'll have to leave it a little bit unknown because i'm filling in 1 plus h as the b value here.

02:23 And then g of 1 is that g of a amount...

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