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.
Graph of Cube Root Function


Properties of Cube Root Function
- Domain = All real numbers
- Range = All real numbers
- 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.3
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
- Compare the graph of p(x) = ∛+5 to the graph of f(x) = ∛.
- Compare the graph of q(x) = ∛-2 to the graph of f(x) = ∛.
- Calculate the average rate of change of r(x) = ∛+2; 4 ≤ x ≤ 8.
Concept Summary


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