Help Please!
How to find the derivative algebraically of f(x)=x^3 at x=-2

Question

isaiah.feynman · Answer

Definition of the derivative. $$\lim_{x \rightarrow a}\frac{ f(a+h)-f(a) }{ x-a }$$

isaiah.feynman · Answer

Where a =-2

anonymous · Answer

So what do I do with the x^3 part?

isaiah.feynman · Answer

Let x = a+h, but don't substitute a =-2 in this part. Substitute it in f(a)

isaiah.feynman · Answer

Hey, sorry don't use that formula. Please use this one its more convenient.$$\lim_{x \rightarrow a}\frac{ f(a)-f(x) }{ x-a }$$

isaiah.feynman · Answer

a=-2. So f(a)=f(-2) and f(x) is x^3.

anonymous · Answer

I'm sorry. I must still be missing something. I know that the correct answer is 12, but I can't seem to get that answer. I keep getting 0.

isaiah.feynman · Answer

Did you use the second formula I gave you?

anonymous · Answer

Yea. It would be (-8)-(x^3)/x^3-(-2), right? I don't know why I'm so confused right now.

isaiah.feynman · Answer

Oh my God, IT'S ALL MY FAULT!!!
What's wrong with me today??!!

anonymous · Answer

Huh?

isaiah.feynman · Answer

I wrote the second formula the wrong way. Its supposed to be $$\lim_{x \rightarrow a} \frac{ f(x)-f(a) }{ x-a }$$

anonymous · Answer

I'm still getting 0 as an answer.

isaiah.feynman · Answer

So to substitute, $$\lim_{x \rightarrow -2}\frac{ x^{3}-(-8) }{ x-(-2) }$$

anonymous · Answer

But then you'll get (-8)-(-8) as the numerator and (-2)-(-2) as the denominator, both equaling 0, wouldn't you?

isaiah.feynman · Answer

No. That expression becomes $$\lim_{x \rightarrow -2}\frac{ x^{3}+8 }{ x+2 }$$
Now solve that equation.

anonymous · Answer

But when you substitute -2 for x, you get (-2)^3, which is -8, and -2. Making it -8+8 over -2+2. Right?

isaiah.feynman · Answer

You DON'T substitute x for -2!!! That would make the denominator zero and division by zero is undefined. That arrow means x approaches -2.

anonymous · Answer

So how will the answer end up being 12?

isaiah.feynman · Answer

I factor the denominator $$\lim_{x \rightarrow -2} \frac{ (x+2)(x^{2}-2x+4) }{ (x+2) }$$.
The x+2's cancel out leaving $$\lim_{x \rightarrow -2} x^{2}-2x+4.$$ Now substitute x for -2 and see what happens.

anonymous · Answer

Oh my goodness, I'm such an idiot. I wasn't expanding x^3. I was keeping it as is. I'm so sorry for all the trouble I put you through. But thank you sooo much! It really is appreciated.

isaiah.feynman · Answer

I made silly mistake too. Lol.