Let ​g(x)equals19 StartRoot x EndRoot plus 2 for xgreater than or equals0. Complete parts​ (a) through​ (d) below.
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Part 1
a. Find the average rate of change of​ g(x) with respect to x over the intervals​ [1,2], [1,1.5], and ​[1,1plus​h].
The average rate of change over the interval​ [1,2] is 
  
7.870058.
​(Round to six decimal places as​ needed.)
Part 2
The average rate of change over the interval​ [1,1.5] is 
select: 23.270155
8.540305.
​(Round to six decimal places as​ needed.)
Part 3
The average rate of change over the interval ​[1,1plus​h] is 
  
StartFraction 19 left parenthesis StartRoot 1 plus h EndRoot minus 1 right parenthesis Over h EndFraction
.
Part 4
b. Make a table of values of the average rate of change of g with respect to x over the interval ​[1,1plus​h] for hequals​0.1, ​0.01, 0.001,​ 0.0001, 0.00001, and 0.000001.

Let's start by finding the average rate of change of \( g(x) = 19\sqrt{x} + 2 \) over the given intervals and then complete the required table of values.

### Part (a): Average Rate of Change
The average rate of change of a function \( g(x) \) over an interval \([a, b]\) is given by:

\[
\text{Average rate of change} = \frac{g(b) - g(a)}{b - a}
\]

1. **Interval \([1, 2]\)**:

\[
g(1) = 19\sqrt{1} + 2 = 19 + 2 = 21
\]

\[
g(2) = 19\sqrt{2} + 2 = 19 \times 1.414213 + 2 \approx 28.870058
\]

\[
\text{Average rate of change} = \frac{g(2) - g(1)}{2 - 1} = \frac{28.870058 - 21}{1} = 7.870058
\]

2. **Interval \([1, 1.5]\)**:

\[
g(1.5) = 19\sqrt{1.5} + 2 = 19 \times 1.224745 + 2 \approx 25.540305
\]

\[
\text{Average rate of change} = \frac{g(1.5) - g(1)}{1.5 - 1} = \frac{25.540305 - 21}{0.5} = 8.540305
\]

So, the correct option is \( 8.540305 \).

3. **Interval \([1, 1 + h]\)**:

\[
g(1 + h) = 19\sqrt{1 + h} + 2
\]

\[
\text{Average rate of change} = \frac{g(1 + h) - g(1)}{h} = \frac{19\sqrt{1 + h} + 2 - 21}{h} = \frac{19(\sqrt{1 + h} - 1)}{h}
\]

### Part (b): Table of Average Rate of Change for Specific Values of \( h \)

We need to calculate the average rate of change for the function over the interval \([1, 1 + h]\) for different small values of \( h \).

\[
\text{Average rate of change} = \frac{19(\sqrt{1 + h} - 1)}{h}
\]

Let's evaluate this expression for the specified values of \( h \).

#### Calculations for \( h \):

1. **\( h = 0.1 \):**

   \[
   \text{Average rate of change} = \frac{19(\sqrt{1.1} - 1)}{0.1}
   \]

2. **\( h = 0.01 \):**

   \[
   \text{Average rate of change} = \frac{19(\sqrt{1.01} - 1)}{0.01}
   \]

3. **\( h = 0.001 \):**

   \[
   \text{Average rate of change} = \frac{19(\sqrt{1.001} - 1)}{0.001}
   \]

4. **\( h = 0.0001 \):**

   \[
   \text{Average rate of change} = \frac{19(\sqrt{1.0001} - 1)}{0.0001}
   \]

5. **\( h = 0.00001 \):**

   \[
   \text{Average rate of change} = \frac{19(\sqrt{1.00001} - 1)}{0.00001}
   \]

6. **\( h = 0.000001 \):**

   \[
   \text{Average rate of change} = \frac{19(\sqrt{1.000001} - 1)}{0.000001}
   \]

Let's calculate these values numerically.Here is the table of average rates of change of \( g(x) \) over the interval \([1, 1 + h]\) for the given values of \( h \):

| \( h \)        | Average Rate of Change                |
|----------------|---------------------------------------|
| 0.1            | 9.273681                              |
| 0.01           | 9.476368                              |
| 0.001          | 9.497626                              |
| 0.0001         | 9.499763                              |
| 0.00001        | 9.499976                              |
| 0.000001       | 9.499998                              |

### Summary:
As \( h \) approaches 0, the average rate of change gets closer to 9.5, which suggests the instantaneous rate of change (or derivative) at \( x = 1 \) is approximately 9.5.

Would you like more details on the calculations, or do you have any questions? 

### Related Questions:
1. What does the average rate of change tell us about the behavior of the function?
2. How does the concept of average rate of change relate to the derivative?
3. Can we find the exact derivative of \( g(x) = 19\sqrt{x} + 2 \)?
4. How would these rates of change change if we considered intervals closer to \( x = 0 \)?
5. What happens to the average rate of change as \( h \) becomes very small?

### Tip:
The average rate of change between two points approximates the slope of the secant line connecting them. As the interval gets smaller, it approaches the derivative at a point.

Let ​g(x) equal 19√x + 2 for x ≥ 0. Complete parts (a) through (d) below. Part 1: 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]. The average rate of change over the interval​ [1,2] is 7.870058 (Round to six decimal places as needed). Part 2: The average rate of change over the interval [1,1.5] is 8.540305. Part 3: The average rate of change over the interval [1,1+h] is (19(√(1+h) - 1)) / h. Part 4: 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 h=​0.1, ​0.01, 0.001,​ 0.0001, 0.00001, and 0.000001.

Learn how to calculate the average rate of change for the function g(x) = 19√x + 2 over various intervals, including [1,2], [1,1.5], and [1,1+h]. This step-by-step solution involves calculus concepts and formulas, suitable for high school students in Grades 11-12.

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