Expert: Robi Bhattacharjee - 3/29/2009

Question
I am trying to figure out what the question is asking for in the following:

Evaluate the functions for the values of x given as 1,2,4,8 and 16. Describe the differences in rate at which each function changes with increasing values of x.

So far this is what i have:

KERRY DAVIS   WEEK 6 ASSIGNMENT 3C
SOUTH UNIVERSITY

Evaluate the functions for the values of x given as 1,2,4,8 and 16. Describe the differences in the rate at which each function changes with increasing values of x:

1:f(x) = 2x -5
if x = 1, then y =  2(1) – 5    or 2 – 5 = -3   (1, -3)
if x = 2, then y = 2(2) – 5   or 4 – 5 = -1   (2, -1)
If x = 4, then y = 2(4) – 5   or 8 – 5 = 3   (4, 3)
if x = 8, then y = 2(8) – 5   or 16 – 5 = 11   (8,11)
If x = 16, then y = 2(16) – 5   or 32 – 5 = 27   (16, 27)
  
The rate of change in y is: -3 to -1 or 2; -1 to 3 or 4; 3 to 11 or 8; and 11 to 27 or 16. Since this is a linear function, the value of x is directly proportional to the value of y. Therefore, if the value of x doubles the value of y doubles as well at a ratio of 2:1 in this function since the slope m = (3 – (-1))/4-2 or 4/2 or 2.
  
2:f(x) = x2 – 4x +3
if x = 1, then y = (1)2 – 4(1) + 3or 1 -4 + 3 = 0    (1,0)
If x = 2, then y = (2)2 – 4(2) + 3 or 4 – 8 + 3 = -1(2, -1)
if x = 4, then y = (4)2 – 4(4) + 3or 16 – 16 + 3 = 3(4, 3)
If x = 8, then y = (8)2 – 4(8)+ 3or 64 – 32 + 3 = 35(8, 35)if x = 16, then y = (16)2 – 4(16)+ 3or 256 -64 +3 = 195      (16, 195)

The rate of change in y is: 0 to -1 or -1; -1 to 3 or 4; 3 to 35 or 32; 35 to 195 or 160. Since this is a quadratic function, the rate of change is related to x. The function is a quadratic function. So the rate change is increased more than the linear function.

     
3:f(x) = x3 + 4x2  - 2x – 3
if x =1, then y = (1)3 + 4(1)2 – 2(1) – 3
or 1 + 4 – 2 – 3 = 0      (1,0)
if x = 2, then y = (2)3 + 4(2)2 – 2(2) – 3
or 8 +16 – 4 – 3 = 17      (2, 17)
if x = 4, then y = (4)3 + 4(4)2 – 2(4) – 3
or 64 + 64 – 8 – 3 = 117      (4, 117)
if x = 8, then y = (8)3 + 4(8)2 – 2(8) – 3
or 512 + 256 – 16 – 3 = 749   (8, 749)
if x = 16, then y = (16)3 + 4(16)2 – 2(16) – 3   or 4096 + 1024 – 32 – 3   = 5085   (16, 5085)

The rate of change in y is: 0 to 17 or 17; 17 to 117 or 100; 117 to 749 or 632; 749 to 508 or 4336. Each increase of x where x doubles is a 0.8 consistent increase of the logarithm of y, i.e. log 17 = 1.2, log 117 = 2.0, log 749 = 2.8 and log 5085 = 3.6. So while the value of x is doubled the value of y increase drastically based upon the degree of x in the polynomial expression, but has a consistent difference of 0.8 when the logarithmic values of y are compared from one point to the next. The function is a third-order polynomial function of x, so the change of f(x) is much higher than the 2nd order quadratic function. The greater the degree in the polynomial as represented by the x n then the greatest increase in the rate of change in y.

4:f(x) = 7x or y = ax then x = ay
If x = 1, then y = 71    or  7x1= 7   (1,7)
If x = 2, then y =72   or 7 X 7 = 49   (2, 49)
If x = 4, then y = 74   or 7X7X7X7 = 2401(4, 2401)
If x = 8, then y = 78    or 7x7x7x7x7x7x7x7 = 5.765 x 10^6      (8, 5.765x10^6)
If x = 16, then y = 716   or 7x7x7x7x7x7x7x7x7x7x7x7x7x7x7x7 =    (16, 3.323x10^13)

The rate of change in y is: 7 to 49 or (7)2; 49 to 2401 or (49)2; 2401 to 5.765x10^6 or (2401)2; and
5.765x10^6 to 3.323x10^13 or (5.765x10^6)2. As each value of x double, the value of y increases by last base given for y then squared as y2, i.e. 72 to 492 to 24012 , and etc.( or 2x = (last value of  y)2 which represents the fastest increase of all the functional expressions given.

5:f(x) = log x
if x = 1, then y = log 1   or  y = 0    (1,0)
if x = 2, then y = log 2   or y = 0.301   (2, 0.3)
if x = 4, then y = log 4   or y = 0.602   (4, 0.6)
if x = 8, then y = log 8   or y = 0.903   (8, 0.9)
if x = 10, then y = log 16 or y = 1.204      (16, 1.2)

The rate of change in y is: 0 to 0.301 or 0.301; 0.301 to 0.602 or 0.301; 0.602 to 0.903 or 0.301 and
0.903 to 1.204 or 0.301. Therefore for every increase in x where x doubles, the increase in y remains
at a consistent rate increase of 0.301. Therefore as x doubles in increase, y increases by log 2 and
represents the slowest rate of increase of all the functional expressions given.

is this correct or am i way off the mark here? I would really appreciate any assistance you can provide.. Thank -you!

Kerry Davis


Answer
Hello Kerry Davis,

Well, I would say that you are right overall, but you have a few problems.

First, if the math class that this homework is for involves calculus, you would have a much easier time. To find the rate of change, you simply find the derivative, so this would be much quicker.

Second, you generally don't say "when x doubles y...". Generally, mathematicians say if x increases by 1, then y increases by... This is the whole idea of slope and rate of change.


I hope this helps,
Robi