# Create two functions that can be developed from y=x^2 and y=x^3 that will give you the instantaneous rate of change without reducing the interval.Yes. The first part of this question asked me to...

Create two functions that can be developed from y=x^2 and y=x^3 that will give you the instantaneous rate of change without reducing the interval.

Yes. The first part of this question asked me to estimate the instantaneous rate of change for the function y=x^2 at x=1, x=2, and x=3 and for the function y=x^3 at x=1, x=2, x=3 which I did and found the rate of change for y=x^2 at x=1 to be 2, x=2 to be 4 and x=3 to be 6.  I also found the rate of change for y=x^3 at x=1 to be 3, x=2 to be 12 and x=3 to be 27.

Then it asked to create two functions that can be developed from the above to give the instantaneous rate of change without reducing the interval.

neela | High School Teacher | (Level 3) Valedictorian

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The instantaneous rate of change of y = x^2 is dy/dx = 2x and y = x^3 separately is dy/dx = 3x^2.

For the first function y = x^2 dy/dx at x= 1, is 2, at x=2 is  dy/dx = 2x =2*2 = 4, and at x= 3,dy/dx = 2x is 2*3 = 6.

The instantaneous rate of change for the second function y = x^3, at x= 1, is dy/dx = 3x^2= 3*1= 3. At x= 2, dy/dx = 3x^2= 3*2^2 = 12. at x= 3, dy/dx = 3*x^2 = 27.

The instantaneous rate for the two functions that are developed from the functions. We consider the following two fuctions:

y = x^2+x^3 and y = x^3/x^2 = x.

The instantaneous rate of change for y= x^2+x^3 is dy/dx= 2x+3x^2. At x= 1, dy/dx = 2+3. At x= 2, dy/dx = 2x+3x^2= 2*2+3*2^2 = 16. At x= 3, dy/dx = 2x+3x^2= 2*3+3*3^2 = 33.

For the function y = x^3/x^2 = x, dy/dx = (x)' = 1. So x' = 1 at x= 1.

At x= 2, dy/dx = (x)' = 1 is the rate of change.

At x= 3, dy/dx = (x)' = 1 is the rate of change.