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.