function help!!?

Question

The function f is defined for the domain -3 f(x) = 9 (x - 1/3)^2 - 11

(i) Find the range of f.

(ii) State the coordinates and nature of the turning point of
(a) the curve y = f(x)
(b) the curve y = If(x)I

Thanks........

Answer 1

the range would be the y coordinates of the extremes (-3 and 3). simply plug them in and calculate. i believe for the curve you will need to find the derivative. after you do that set it equal to zero and solve to find the turning point. the second derivative will tell you if it is concave up or down. thats what i remember (havent had calculus in a year or so).

Answer 2

The vertex of the parabola is at (1/3,-11) and is a minimum
The range is -11 to 89
If you take I f I at the points where f=0 the derivative does not exist.
9(x-1/3)^2-11=0
x= 1/3+-sqrt11/3 = 1/3(1+-sqrt11)
At those points I f I has a minimum (without derivative) and at x=1/3 we have a maximum with derivative =0
The range now is 0 to 89

Answer 3

i. The maximum must be at f(-3) = 89. The minimum will be at f(1/3) = -11. The range is (-11,89).

iia. It is the minimum of a parabola at (1/3, -11). This is confirmed by differentiating the function and equating to zero.
f' = 18x - 6 = 0 -> x = 1/3

iib. There are three turning points. One is a parabola maximum at (1/3,11). The other two are minima at f(0) = 0 and at x = -0.772 and 1.44.