Q: Let f(x) = 3x² – x
a. Find the average rate of change of f(x) with the respect to x as x changes from x = 0 to x = 1/16
b. Use calculus to find the instantaneous rate of change of f(x) at x=0 and compare with the average rate found in part (a).
A: Let’s start with part (a). The average rate of change is nothing more than a slope of a line. Remember,
Slope = m = (y2 – y1) / (x2 – x1)
Well, we know x1 = 0 and x2 = 1/16… How do we find y1 and y2? We plug each x into the original function to find the corresponding y value, like so:
f(x) = 3x² – x
f(0) = 3(0)² – 0 = 0
So, the point (x1 , y1) = (0, 0)
Now try for x = 1/16:
f(1/16) = 3(1/16)² – 1/16 = 3(1/256) – 1/16
= 3/256 – 1/16 = -13/256
So, the point (x2 , y2) = (1/16, -13/256)
Now, calculate the average rate of change (slope):
m = (y2 – y1) / (x2 – x1) = (-13/256 – 0) / (1/16 – 0) = (-13/256) / (1/16)
[remember, diving by a fraction is the same as multiplying by its reciprocal!]
= -13/256 * 16/1 = -13/16
The average rate of change from x = 0 to x = 1/16 is -13/16
(b) Now, the instantaneous rate of change is asking for the rate of change at a single point. In our case, the specific point is when x = 0. We need calculus to find the rate of change at a single point….. Remember, the derivative gives us the rate of change (slope) at a single point.
So, take the derivative first:
f(x) = 3x² – x
f ‘ (x) = 6x – 1
Now, we are curious about the point where x = 0… So, plug in 0 for x:
f ‘ (x) = 6(0) – 1 = -1
The instantaneous rate of change of the function when x = 0 is -1.
How does this answer compare the the answer in part (a)? Well, an average rate of change of -13/16 is pretty darn close to the instantaneous rate of change which was -1. The average rate of change gives a pretty good estimate of the instantaneous rate of change, as long as you are on a continuous function and use two points that are pretty close together.
A very good explanation for someone like myself who has just started this topic at school.
LikeLiked by 1 person