Rate of change word question

[jgarcia]

Rate of change word question (differentiation)

Here is a question that I am struggling with. I am not sure what it is asking me to do, bet here goes:

Let f(x)=x(1 – 2x)

a) Find the average rate of change of f(x) with respect to x as x changes from x = 0 to x = ¹/₂.

b) Use calculus to find the instantaneous rate of change from f(x) at x = 0 and compare with average found in part (a).

Here is how I've approached the problem:

dy/dx= average rate of change f(x)

so I find for y from the derivative of f(x)=x(1 – 2x) or f'(x)=x(1 – 2x) or y'=x(1 – 2x)

y'=x(1 – 2x)
y'=x – 2x²
y'=1– 4x

m=1/4

so then I solve for y at 0 ...

y=x(1 – 2x)
y=(0)(1 – 2(0))
y=(0)(1 – 0)
y=0

so then I solve for y at 1/2 ...

y=x(1 – 2x)
y=(1/2)(1 – 2(1/2))
y=(1/2)(1 – 1)
y=0

y-y₁=m(x-x₁₎
y-0=m(0-1/<sub>2)
y=-1/2

Rate of change -1/2</sub>

 

[stapel]

Let f(x)=x(1 – 2x)

a) Find the average rate of change of f(x) with respect to x as x changes from x = 0 to x = ¹/₂.

b) Use calculus to find the instantaneous rate of change from f(x) at x = 0 and compare with average found in part (a).

Here is how I've approached the problem:

dy/dx= average rate of change f(x)

No; by definition, the derivative is the instantaneous change. The average rate of change is found in the same way you did back in algebra; namely, by finding the values at the endpoints of the interval, subtracting to find the absolute change, and then dividing by the length of the interval.