Find the average rate of change of f ( x ) = 4 x 2 − 2 on the interval [ 5 , b ] . Your answer will be an expression involving b

Answer 1

f(x) = 4x^2 - 2

f(b) = 4b^2 - 2

f(5) = 4(5)^2 - 2 = 4(25)-2 = 100-2 = 98

average rate of change = [f(b)-f(5)]/((b-5) = (4b^2-2 -98)/(b-5) = (4b^2-100)/(b-5)

for f(x) =4x^2 -2

the instantaneous rate of change is the derivative f'(x)=8x, if b=5 then f''(5) = 8(5) = 40

that's the slope of the tangent line at x=5

if b doesn't equal 5, then the slope of the secant line is the average rate of change.

the slope of the secant line approaches the slope of the tangent line as b approaches 5

It's similar to a speed limit sign. Trick question is it average or instantaneous speed that's a traffic infraction? It's instantaneous speed that's illegal. but the rate of change difference with functions is the same for average vs. instantaneous.

Police pull you over and you explain you've been driving for one hour and only gone 55 miles in a 55mph zone, so you complied. But that's just an average speed. If at any time you go above 55, even for one instant and they get you on radar, you've got a ticket.