What is the average rate of change of f(x) = (x + 3)2 − 2

When we have a function f(x), the average rate of change in the interval (a, b) is:

In this case, we have the function:

f(x) = (x + 3)^2 - 2

(but we do not have the interval, and I couldn't find the complete question online)

So if for example, we have the interval (2, 4)

The average rate of change will be:

If instead, we want the rate of change in a differential dx around the value x, we need to differentiate the function (this is way more complex, so I will define some rules first).

Such that the rate of change, in this case, will be:

f'(x) = df/dx

For a function like:

g(x) = x^n + c

g'(x) = n*x^(n - 1)

And for:

h(x) = k( g(x))

h'(x) = k'(g(x))*g'(x)

So here we can write our function as:

f(x) = k(g(x)) = (x + 3)^2 - 2

where:

g(x) = x + 3

k(x) = x^2 - 2

Then:

f'(x) = 2*(x + 3)*1 = 2*x + 6

That is the rate of change as a function of x (but is not an "average" rate of change)

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

what is the question? i am guessing you are looking for the area.

to find an area, you have to do length x width.

in this case, it would be 12 ft x 4 ft = 48 sq. ft