Is My AP Calculus Solution Correct?

[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.

Thanks in advance!

 

[HallsofIvy]

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

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?

b. f'(x) = 3x$$^{2}$$ - 7
m = f'(-1) = 3(-1)$$^{3}$$ - 7 = -4

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

f'(-1) = (-1)$$^{3}$$ - 7(-1) + 6 = 12

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

Final Answer: y - 12 = -4(x+1)

Try again!

c. f(b) = f(3) = 3$$^{3}$$ - 7(3) + 6 = -6

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

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.

Thanks in advance!

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

 

[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.

 

[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.

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

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!

 

[HallsofIvy]

It is never rude to point out a critical error!