# Re: Well then...

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Posted by Joel on October 16, 2002 at 19:21:18:

In Reply to: Well then... posted by Subhotosh Khan on October 16, 2002 at 18:38:26:

: : No, we are not quite there yet...I believe we study that in the next unit.

: :
: : : : Find the slope of the tangent line to the curve y = x^3 at the point (-1, -1).

: : : : So we're using...

: : : : lim x-> -1

: : : : [ f(x) - f(a) ] / x - a

: : : : lim x-> -1

: : : : [x^3 - f(-1)] / x - (-1) = [x^3 - (-1)] / x - (-1)

: : : : = lim x->-1 [x^3 + 1] / x + 1

: : : : This is where I'm stuck...can anyone shed some light on what my next step(s) is? Thanks.

: : : :
: *************************************
: You need to find following first:

: lim(h - > 0)[{f(x+h)-f(x)}/h]

: = lim(h - > 0)[{(x+h)^3-(x)^3}/h]

: = lim(h - > 0)[{3*x^2*h + 3* x*h^2 + h^3}/h]

: You can evaluate that limit and you will get another function of 'x' - say f'(x). That is the slope of the tangent line at 'x'.

: Now find f'(-1) to get the answer you want....

They may not yet have learned to transform x->a to h->0. That's the next lesson.

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