# AP Style Problem - Derivatives

1. Oct 8, 2007

### yokoloko13

Hey guys. These are the first AP Problems I'm doing. Here is all my work.

Let f be the function given by f(x) = x$$^{3}$$-7x+6.

a. Find the zeros of f.
b. Write an equation of the line tangent to the graph of f at x = -1
c. Find the number c that satisfies the Mean Value Theorem for f on the closed interval [1,3]

So what I did is:
a. 0 = x$$^{3}$$ - 7x + 6
-6 = x(x$$^{2}$$-7)
Final Answer: x = +/-1, 6

b. f'(x) = 3x$$^{2}$$ - 7
m = f'(-1) = 3(-1)$$^{3}$$ - 7 = -4
f'(-1) = (-1)$$^{3}$$ - 7(-1) + 6 = 12
Final Answer: y - 12 = -4(x+1)

c. f(b) = f(3) = 3$$^{3}$$ - 7(3) + 6 = -6
f(a) = f(1) = 1$$^{3}$$ - 7(1) + 6 = 0
f'(c) = (f(b)-f(a)) / (b-a) = (-6 - 0) / (3-1) = -3
-3 = 3c$$^{2}$$ - 7
Final Answer: c = 2$$\sqrt{3}$$ / 3

Please tell me what is right and wrong? I am fairly confident on a and b, but shaky on c because I haven't studied that yet and don't know if it's right at all.

2. Oct 8, 2007

### HallsofIvy

I don't understand where you would get +/- 1, 6 from "$-6= x(x^2- 7)$ or even how that last form helps. Even if you don't know how to solve this equation, you should be able to check: If x= 1, then $1^3- 7(1)+ 6= 0$ so that is a solution. If x= -1, $(-1)^3- 7(-1)+ 6= -2$ not 0. Finally, if x= 6, $6^3- 7(6)+ 6= 216- 42+ 6= 180$, nowhere near 0!. Since 1 is a root, you can divide $x^3- 7x+ 6$ by x-1 and find that $x^3- 7x+ 6= (x-1)(x^2+ x- 6)$. what are the other two roots?

you mean $3(-1)^2- 7= -4$ but yes, that is the correct slope.

Well, you mean f(-1), not f'(-1), but even then, $(-1)^3- 7(-1)+ 6= -1-7+ 6= -2$ not 12!

Try again!

?? $3^3 -7(3)+ 6= 27- 21+ 6= 12$!

Your calculus is good but your arithmetic is atrocious! Perhaps you are trying to do it too fast.

3. Oct 8, 2007

### yokoloko13

Lol thanks, I guess. Yes, I was in a rush to leave. Thank you for your help. I will go back and check everything. Hectic day, my friend. Forgive me for the careless mistakes.

4. Oct 9, 2007

### yokoloko13

So I went back and reworked it.
For a) I did the synthetic division and got "x = 1,2, -3" for my final answer.

For b) Don't want to be rude, but my answer was correct. Your mistake was in the underlined part (-1)^3 -7(-1)+ 6= -1-(-7)+ 6= 12. Subtracting a negative (when you just subtracted the positive) or how ever way you would like to look at it.

c) Yes, my math was horrible there.
f(b) = f(3) = 3$$^{3}$$ - 7(3) + 6 = 12
f(a) = f(1) = 1$$^{3}$$ - 7(1) + 6 = 0
f'(c) = (f(b)-f(a)) / (b-a) = (12 - 0) / (3-1) = 6
6 = 3c$$^{2}$$ - 7
c = $$\sqrt{13/3}$$

I'm quite sure it's fixed now! Thanks for all the help!

5. Oct 10, 2007

### HallsofIvy

It is never rude to point out a critical error!