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?

2 Answers
Jan 16, 2018

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#.

Jan 16, 2018

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#