slope of tangent line on a curve?

**yoman360** · August 20th 2009, 09:44 PM

Find the slope of the tangent to the curve $f(x) = x^3-4x+1$ at the point where x=a. This is what I get
$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$
$\lim_{x\to a}\frac{(x^3-4x+1)-(a^3-4a+1)}{x-a}$
$\lim_{x\to a}\frac{(x^3-4x+1)+(-a^3+4a-1)}{x-a}$
$\lim_{x\to a}\frac{x^3-4x-a^3+4a}{x-a}$

Then I'm stuck from there. Oh and don't use derivatives or that $\frac{dy}{dx}$ stuff because we haven't learned that yet.

**earboth** · August 20th 2009, 10:39 PM

Originally Posted by yoman360

Find the slope of the tangent to the curve $f(x) = x^3-4x+1$ at the point where x=a. This is what I get
$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$
$\lim_{x\to a}\frac{(x^3-4x+1)-(a^3-4a+1)}{x-a}$
$\lim_{x\to a}\frac{(x^3-4x+1)+(-a^3+4a-1)}{x-a}$
$\lim_{x\to a}\frac{x^3-4x-a^3+4a}{x-a}$

Then I'm stuck from there. Oh and don't use derivatives or that $\frac{dy}{dx}$ stuff because we haven't learned that yet.

I take your last term:

$\lim_{x\to a}\frac{x^3-4x-a^3+4a}{x-a} = \lim_{x\to a}\frac{x^3-a^3 - 4x+4a}{x-a}$

${\color{white}\lim_{x\to a}\frac{x^3-4x-a^3+4a}{x-a}} = \lim_{x\to a}\frac{(x^3-a^3) - 4(x-a)}{x-a}$

${\color{white}\lim_{x\to a}\frac{x^3-4x-a^3+4a}{x-a}} = \lim_{x\to a}\frac{((x-a)(x^2+ax+a^2)) - 4(x-a)}{x-a}$

Now cancel (x-a):

${\color{white}\lim_{x\to a}\frac{x^3-4x-a^3+4a}{x-a}} = \lim_{x\to a}(x^2+ax+a^2 - 4) = \boxed{3a^2-4}$

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