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April 1, 2015

April 1, 2015

Posted by **Yoona** on Monday, February 13, 2012 at 7:24pm.

(i) f''(x)=24x-18

(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)

- calculus -
**Reiny**, Monday, February 13, 2012 at 7:45pmf'(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.

- calculus -
**Parker**, Monday, April 15, 2013 at 9:40pmf''(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(2x-3) = 0

x = 0 or x = 3/2

Just evaluate f(0) and f(3/2) to find the y-coordinates.

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