by the time you get to calculus, you certainly should know how to plug in a value and evaluate a function! Where do you get stuck?
(a) q(5) = 10300*e^(-0.34*5) = 1882
(b) as you know, if
y = e^u
y' = e^u u'
And, as you know, the derivative is the rate of change.
So, q' = 10300*e^(-0.34p)(-.34) = -3502e^(-.34p)
(c) now plug in p=5 to get q'(5)
extra credit: what does it mean that q'(p) is negative?
q = f(p) = 10300e^−0.34p
a) Find the number of products sold when the price of the product is $5. (Round your answer to the nearest whole number.)
Number of products sold:
b) Find a formula for the rate of change in the number of products sold when the price is p dollars.
f�Œ(p) =
c) What is the rate of change in the number of products sold when the price is $5? (Round your answer to one decimal place.)
Rate of Change:
2 answers
satisfy but not enough
1 answer