Question

Consider the function `f(x) = sqrt(1-x^2)` on the interval [0, -1], how do you find the average or mean slope of the function on this interval?

Answer 1

The slope of a function `f(x)` at any point is equal to `d/dx(f(x))`.

The average value of a function `g(x)` from `x in [a,b]` is given by `1/(b-a)int_a^bg(x)\ dx`.

We want to find the average slope of the function `f(x)=sqrt(1-x^2)` on `x in [-1,0]`.

Its slope at any given point is `d/dx(sqrt(1-x^2))`. Thus, the average slope from `x=-1` to `x=0` is given by
`\ \ \ \ \ \ 1/(0-(-1))int_-1^0d/dx(sqrt(1-x^2))\ dx`

By the Fundamental Theorem of Calculus, we know that integration and differentiation are inverse operations. Thus, we have the above expression equal to
`=[sqrt(1-x^2)]_-1^0`
`=1`

We can verify this via a graph:
graph{sqrt(1-x^2) [-3,3,-1.5,1.5]}

By inspection, it seems that the average slope of `x in [-1,0]` is `1`.

Answer 2

The slope is the same as the rate of change.

Explanation:

Note that the "interval `[0,-1]` should mean the set of all `x` with `0 <= x` and `x <= -1`. There is no such `x`, so the "interval" is the empty set.
I will assume that you mean `[-1,0]`.

The slope is the same as the rate of change.
So the average slope on `[-1,0]` is equal to the average rate of change on the interval:

`(f(0)-f(-1))/(0-(-1)) = (sqrt1 - sqrt0)/1 = 1`

If the intended interval is `[0,1]`, then the average slope is

`(f(1)-f(0))/(1-0) = (sqrt0 - sqrt1)/1 = -1`