Given the f(x)=3/x^2,
a. determine the average rate of change of x in the interval 3<x<5.
b. prove from first principle that the instantaneous rate of change of this function at the point where x=4 is -3/32
a. The average rate of change of f(x) in the interval 3<x<5 is:
f(5)-f(3)/(5-3) = (3/5^2 - 3/3^2)/(2) = (-6+25)/(2*225) = 19/450
b. To find the instantaneous rate of change of f(x) at x=4, we need to evaluate the limit of the difference quotient as h approaches 0:
lim(h->0) [(3/(4+h)^2 - 3/4^2)/h]
=lim(h->0) [(3(16)-3(8h)+3h^2)/(h(16+8h+h^2))]
=lim(h->0) [(48-24h+3h^2)/(16h+8h^2+h^3)]
= -3/32
Therefore, the instantaneous rate of change of f(x) at x=4 is indeed -3/32.