calculus
posted by Yoona .
1. Let f be the function that is defined for all real numbers x and that has the following properties.
(i) f''(x)=24x18
(ii) f'(1)=6
(iii) f'(2)=0
a. Find each x such that the line tangent to the graph of f at (x, f (x)) is horizontal
b. Write an expression for f (x)

f'(x) = 12x^2  18x + c
for f'(1) = 6
12  18 + c = 6
c = 0
so f'(x) = 12x^2  18x
Then you say f'(2) = 0 , which does not verify with what I have so far.
Did you mean f(2) = 0 ?
I will assume you meant that.
f(x) = 4x^3  9x^2 + k
given f(2) = 0
32  36 + k = 0
k = 4
f(x) = 4x^3  9x^2 + 4
leaving up to you: .....
a) .......
tell me what you would do. 
f''(x) = 24x  18
f'(x) = 12x^2  18x + C
Since f'(1) = 6, then
6 = 12(1)^2  18(1) + C
6 = 12  18 + C
6 = 6 + C
C = 0
So f'(x) = 12x^2  18x
f(x) = 4x^3  9x^2 + C
Since f(2) = 0, then
0 = 4(2)^3  9(2)^2 + C
0 = 4(8)  9(4) + C
0 = 32  36 + C
0 = 4 + C
4 = C
f(x) = 4x^3  9x^2 + 4
For Part A, you know that f'(x) = 12x^2  18x
f'(x) = 0
12x^2  18x = 0
6x(2x3) = 0
x = 0 or x = 3/2
Just evaluate f(0) and f(3/2) to find the ycoordinates.
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