Question

The function ​f(x)equalsxsquared describes the area of a​ square, f(x), in square​ inches, whose sides each measure x inches. a. nbsp If x is​ changing, find the average rate of change of the area with respect to x as x changes from 5 inches to 5.2 inches and from 5 inches to 5.02 inches. b. nbsp If x is​ changing, find the instantaneous rate of change of the area with respect to x at the moment when xequals5 inches.

Answer 1

A) As x changes from 5 inches to 5.2 inches , average rate of change = 10.2π in²/in

- As x changes from 5 inches to 5.02 inches , average rate of change = 10.02π in²/in

B) Instantaneous rate at x=5 is; 10π in²/in

Explanation:

A) We are given that the function which is the area of the square is given by:

f(x) = πx²

As x changes from 5 to 5.2, a=5 and h = 0.2

Average rate of change between 2 points A and B is given by;

[f(a+h) - f(a)]/h

Thus, at a=5 and h = 0.2, we have;

[f(5+0.2) - f(5)]/0.2

= [f(5.2) - f(5)]/0.2

Since, f(x) = πx², let's substitute for x to obtain ;

[(π•5.2²) - (π•5²)]/0.2 = 10.2π in²/in

As x changes from 5 to 5.02, a=5 and h = 0.02

Average rate of change between 2 points A and B is given by;

[f(a+h) - f(a)]/h

Thus, at a=5 and h = 0.02, we have;

[f(5+0.02) - f(5)]/0.02

= [f(5.02) - f(5)]/0.02

Since, f(x) = πx², let's substitute for x to obtain ;

[(π•5.02²) - (π•5²)]/0.02 = 10.02π in²/in

B) The instantaneous rate of change of the area with respect to x will be the derivative of f(x).

Thus, since f(x) = πx²

f'(x) = 2πx

So, at x = 5 inches,

Instantaneous rate = 2π(5) = 10π

Answer 2

a. 10.2 in²/in

10.02 in²/in

b. 10 in²/in

Explanation:

The area of a square is given by the following function:

`f(x)=x^2`

The average rate of change between two points a and b is:

`r = (f(b)-f(a))/(b-a)`

a. For x = 5 and 5.2:

`r = (5.2^2-5^2)/(5.2-5) \\r=10.2\ in^2/in`

For x = 5 and 5.02

`r = (5.02^2-5^2)/(5.02-5) \\r=10.02\ in^2/in`

b. The instantaneous rate of change in area with respect to x is given by:

`f'(x)=2x`

For x = 5

`f'(5) =2*5 = 10\ in^2/in`