PRECACULUS 12 GRADE

Question

F(X)= 5X^2-11X-9, FIND THE AVERAGE RATE FROM a=2 to a=2+h i find average 5h+9 , but the says it incorrect. WHAT IS WRONG, I DID NOT UNDERSTAND

Answer 1

average rate of change = secant line slope, if you graph it = the difference quotient = (f(2+h)-f(2))/(2+h-2)

f(x) = 5x^2 -11x -9

f(2) = 20-22-9 = -11

f(2+h) = 5(2+h)^2 -11(2+h) -9

= 5(4+4h +h^2) -22 -11h -9

= 20+20h +5h^2 -31 -11h

= -11+5h^2 +9h

f(2+h)-f(2) = 5h^2 +9h -11-(11) = 5h^2+9h

(f(2+h)-f(2))/h = 5h +9

the instantaneous rate of change at x=2 is the slope of the tangent line = derivative = f'(x) = 10x-11

f(2) = 20-11= 9 which is the same as the slope of the secant line when h=0 or rather as h approaches zero as a limit

Answer 2

[F(2 + h) - F(2)]/h = [5(2 + h)² - 11(2 + h) - 9 - (5·2² - 11·2 - 9)]/h = (20 + 20h + 5h² - 22 - 11h - 9 + 11)/h = 5h + 9