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### Evaluate the Square Root and Cube Root:

1. Evaluate each expression:

• √25
• √64
• ∛27
• ∛−64

Solution:

• √25=5
• √64=8
• ∛27=3
• ∛−64=−4

2. Explain how ∛−27 =−3?

Solution:

(−3)3=−27,

So, ∛-27=−3

### Cube Root Function

The function f(x) = ∛x is the cube root function.

### Properties of Cube Root Function

1. Domain = All real numbers
2. Range = All real numbers
3. For f(x) = -∛x, the x–intercept and y–intercept of the graph of the function are both 0.

Note: The graph is increasing for all values in the domain of f.

Example:

What are the maximum and minimum values for f(x) = ∛x over the interval −27≤x≤27?

Solution:

The maximum value for f(x) = ∛x when −27≤x≤27 is 3.

The minimum value for f(x) = ∛x when −27≤x≤27 is -3.

Since the function is always increasing, the maximum and minimum values of the function occur at the endpoints of the given interval.

### Translation of Cube Root Function

Example:

The graph of g(x) = ∛x+4 compared to the graph of f(x) = ∛x

Solution:

The graph of g(x) = ∛x+4 is vertical translation of f(x) = ∛x.

When constant is added to output of the cube root function f(x) = ∛x, the graph of resulting function,

g(x) = ∛x+k, is vertical translation of the graph of f(x).

The domain and range for both the functions are all real numbers

Example:

The graph of g(x) = ∛x+6 compared to the graph of f(x) = ∛x.

Solution:

The graph of g(x) = ∛x+6 is horizonal translation of f(x) = ∛x.

When constant is subtracted from input of the cube root function f(x) = ∛x. , the graph of resulting function, is horizontal translation of the graph of f. The domain and range for both the functions are all real numbers.

#### Model a Problem Using the Cube Root Function

Example:

An original clay cube contains 8 in.3 of clay. Assume that the new package will be a cube with volume x in.3. For what increases in volume would the side length increase between 1 in. and 2 in.?

Solution:

Let the volume of new package is x in.3

And the volume of the old package is 8 in.3

The change in side length of the cube is

f(x) = ∛x-8

Graph of (x) = ∛x-8:

From the graph it shows that f(9)=1 and f(16)=2

So, for increases in volume between 9 and 16 in.

the side length would increase by 1 to 2 in.

### Rate of change of square root function

Example:

For the function f(x) = ∛x-2. , how does the average rate of change from x=2 to x=4 compared to the average rate of change from x=4 to x=6?

Solution:

Step 1:

Evaluate the function for the x – values that correspond to the endpoints of each interval.

Interval: 2≤x≤4

f(2) = ∛2−2 = ∛0 =0

f(4) = ∛4−2 = ∛2 ≈1.25

Interval: 4≤x≤6

f(4) = ∛4−2 = ∛2 ≈1.25

f(6) = ∛6−2 = ∛4 ≈1.58

Step 2:

Find the average rate of change over each interval

From x=2 to x=4:

f(4)−f(2)/4−2 ≈ 1.25−0/4−2 =1.25/2 ≈ 0.625

From x=4 to x=6

f(6)−f(4)/6−4 ≈ 1.58−1.256/6−4 =0.33/2 ≈ 0.165

The average rate of change of the function f(x) = ∛x−2 appears to decrease when x≥2 and as the x-values corresponding to the endpoints of the interval increase. This is consistent with the curve becoming less steep when x≥2 and x increases.

Example:

Which function has the greater average rate of change over the interval 0≤x≤5: The translation of f(x) =∛x to the right 1 unit and up 2 units, or the function g(x) = ∛x−2.

Solution:

The translation of f(x) = ∛x to the right 1 unit and up 2 units is h(x) = ∛x−1+2

h(0)= ∛0-1+2 = ∛−1+2 = −1+2 = 1

h(5)= ∛5-1+2 = ∛4+2 ≈ 1.58+2 = 3.58

Average rate of change = h(5)−h(0)/5−0 ≈ 3.58−1/5=2.58/5 ≈ 0.516

Given function

g(x) = ∛−2

The average rate of change of function g(x) over the interval 0 ≤ x ≤ 5

g(0)= ∛0+2 = 0+2 =2

g(5)= ∛5+2 ≈ 1.709+2 = 3.709

Average rate of change = g(5)−g(0)/5−0 ≈ 3.709−2/5 = 1.7095 ≈ 0.341

Conclusion:

The average rate of change of function h(x) is greater than the average rate of change of function g(x)  over the interval 0 ≤ x ≤ 5.

#### Exercise

1. Compare the graph of p(x) = ∛+5 to the graph of f(x) = ∛.
2. Compare the graph of q(x) = ∛-2 to the graph of f(x) = ∛.
3. Calculate the average rate of change of r(x) = ∛+2; 4 ≤ x ≤ 8.

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