Evaluate the derivative of $x^3 - 3x +1$ using the $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ definition to find the tangent of the curve at the point $(2, 3)$.

I already calculated this derivative using $ x = a + h$ for the above mentioned definition and this is what I got, for what it's worth:

$$\lim_{h \to 0} \frac{((2+h)^3 - 3(2+h) +1) - 3)}{h}$$

$$=\lim_{h \to 0} \frac{h(9 + 6h)}{h} = 9$$

The problem I encounter when using the $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ definition is that I can't factor the numerator of $\lim_{x \to 2} \frac{x^3 - 3x +1 - 3}{x - 2}$ to get rid of that $(x-2)$ in the denominator.

Can anybody give me a hint as to what trickery I can use to factor that numerator?

  • 2
    $\begingroup$ Use polynomial division $\endgroup$ – rlartiga Jul 30 '14 at 14:10
  • $\begingroup$ $$x^3-3x-2=(x-2)(x^2+2x+1)$$ $\endgroup$ – rlartiga Jul 30 '14 at 14:11
  • $\begingroup$ I understand now, thanks, rlartiga. $\endgroup$ – Kermit the Hermit Jul 30 '14 at 15:17

Note that $$x^3 - 3x - 2 = (x - 2)(x^2 + 2x + 1) = (x-2)(x+1)^2$$

If all else fails when you are unable to see the factorization, you can use polynomial long division: Dividing, in this case, by $x-2$ yields the quadratic factor $(x^2 + 2x + 1)$.

  • $\begingroup$ I have never really done polynomial long division before, but I did some reading and examples and now I can appreciate your answer. Did you come up with that factorization from the top of your head? I can now see how that factorization can come from polynomial long division, but not from just looking at it. $\endgroup$ – Kermit the Hermit Jul 30 '14 at 15:14
  • $\begingroup$ I saw it by inspection, but that's because I've factored a massive number of polynomials! Look up the "rational root theorem". That can help narrow the possible factors, provided the roots are rational. Otherwise, if I can't eyeball, polynomial long division is the route to go. $\endgroup$ – Namaste Jul 30 '14 at 15:27
  • $\begingroup$ That's a cool trait that you've acquired, I'm envious. I hate asking you another question, but you seem really knowledgeable and like they say, make hay while the sun shines. Is there any book you can recommend me that will introduce me to most of the 'essential' algebra techniques. I ask since it seems most of my biggest obstacles when trying to teach myself calculus is the algebra side of it. $\endgroup$ – Kermit the Hermit Jul 30 '14 at 15:56
  • $\begingroup$ Have you checked out Khan's Academy (It's on-line, free, with video tutorials and exercises, etc. You'll see menus when you scroll through: math, algebra, <topic>, and can progress at your own speed. Paul's Math Notes (online) are good, too. You might want to also search this site for "reference-request, algebra-precalculus" to check out users' questions that have been asked on this topic, and the many answers. $\endgroup$ – Namaste Jul 30 '14 at 15:59
  • $\begingroup$ That's a wealth of information. You're a scholar and a gentleman, thanks, amWhy. $\endgroup$ – Kermit the Hermit Jul 30 '14 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.