Answer:
Intercepts:
x = 0, y = 0
x = 1.77, y = 0
x = 2.51, y = 0
Critical points:
x = 1.25, y = 4
x = 2.17
, y = -4
x = 2.8, y = 4
Inflection points:
x = 0.81, y = 2.44
x = 1.81, y = -0.54
x = 2.52, y = 0.27
Step-by-step explanation:
We can find the intercept by setting f(x) = 0


where n = 0, 1, 2,3, 4, 5,...

Since we are restricting x between 0 and 3 we can stop at n = 2
So the function f(x) intercepts at y = 0 and x:
x = 0
x = 1.77
x = 2.51
The critical points occur at the first derivative = 0


or

where n = 0, 1, 2, 3

Since we are restricting x between 0 and 3 we can stop at n = 2
So our critical points are at
x = 1.25, 
x = 2.17
, 
x = 2.8, 
For the inflection point, we can take the 2nd derivative and set it to 0



We can solve this numerically to get the inflection points are at
x = 0.81, 
x = 1.81, 
x = 2.52, 
Try adding 11, 11 times. It is a long process, but it should give you what you are looking for. First. to break it down add 10 eleven time, then after you get 110 add the remaining 11.
1. y = 4x - 3
2. y = -x + 4
3. y = 1/3x +1
4. y = ----
5. y = 1/7x + 2
6. y = -4x + 2.75
7. y = 3x + 5
8.
9.
10.
11.
12.
Answer:
The frequency distribution is given below:
Class Frequency
0 - 9 9
10 - 19 6
20 - 29 3
30 - 39 3
40 - 49 3
From the given 4 options, the histogram in option A accurately displays this data. Hence the option A is correct.