given f(x)=5x^2-5x+3, find the average rate of change of f(x) on the interval [4,4+h]. Your answer will be an expression involving h.

Answer 1

f(x) = 5x^2 -5x +3

f(4) = 80-20+3 = 63

f(4+h) = 5(4+h)^2 -5(4+h)+3 = 80+40h+5h^2 -20-5h +3 = 63+35h+5h^2

(f(4+h)-f(4))/(4+h-4) = (63+35h+5h^2 -63)/h = (5h^2+35h)/h = 5h+35

if done right you should get f'(4) = 10(4)-5 = 35 when h=0, which is the derivative of f(x) evaluated at x=4

it worked, so 5h+35 is correct

the average rate of change = the instantaneous rate of change when h becomes zero

Answer 2

Average rate of change = (f(b) - f(a)) / (b - a)

f(4+h) = 5 * (4+h)^2 - 5*(4+h) +3 = 5 (16 + 8h + h^2) - 5(4+h) +3 = 80 + 40h +5h^2 - 20 - 5h +3

= 5h^2 + 35h + 63

f(4) = 5*4^2 - 5*4 + 3 = 80 - 20 + 3 = 63

Now, average rate of change = (f(b) - f(a)) / (b - a) = ((5h^2 + 35h + 63) - 63) / ( 4+h - 4)

= (5h^2 + 35h) / h

= 5h + 35