Thread: average rate of change

for this problem, can you use either equations?

f(x)-f(c)
x-c

or

m= (y1-y2)/(x1-x2)

or

f(b)-f(a)/b-a

anyway the question is compute the average rate of change of function f(x)=1/x-2 on (0,1)= x: x is greater than or equal to 0 and less than or equal to 1.
And find the instantaneous rate of change of f at x=0 and x=1

the average rate of change of the function on the closed interval is

Originally Posted by phalange

for this problem, can you use either equations?

f(x)-f(c)
x-c

or

m= (y1-y2)/(x1-x2)

or

f(b)-f(a)/b-a

I hope you mean (f(b)- f(a))/(b-a) (note parentheses like you used in the "m" equation).

anyway the question is compute the average rate of change of function f(x)=1/x-2 on (0,1)= x: x is greater than or equal to 0 and less than or equal to 1.
And find the instantaneous rate of change of f at x=0 and x=1

Yes, with the correct parentheses, all of those are equivalent to the average rate of change. The "instantaneous rate of change" is the derivative or, equivalently, the limit of those "average rate of change" formulas.