Manipulating LimitsDate: 10/13/2003 at 21:45:29 From: Dijana Subject: calculus I have a three-part question: (a) If g(x)=x^(2/3), show that g'(0) does not exist. (b) If a is not equal to 0, find g'(a). (c) Show that y=x^(2/3) has a vertical tangent line at (0,0). I used the formula f'(a) = lim x=>a where f(x)-f(a)/(x-a), but I don't understand what to do with the question. Date: 10/14/2003 at 07:10:21 From: Doctor Luis Subject: Re: calculus Hi Dijana, If we look carefully at x^(2/3) - a^(2/3) f'(a) = lim -------------------- x->a x - a we see that we can't just evaluate it at x=a because we get division by zero. So, we must find a way to make the top cancel out the bottom. First of all, the numerator looks like a difference of squares. That seems to be a big hint. x^(2/3) - a^(2/3) = (x^(1/3) - a^(1/3)) * (x^(1/3) + a^(1/3)) So, we have to get cube roots out of the denominator somehow. We can do this by writing x as (x^(1/3))^3 and 'a' as (a^(1/3))^3. This means that x-a can be written as a difference of cubes: x - a = (x^(1/3))^3 - (a^(1/3))^3 And what do we know about differences of cubes? Well, there's the identity m^3 - n^3 = (m - n) * (m^2 + m*n + n^2) which means we can rewrite x-a as x - a = (x^(1/3) - a^(1/3)) * (x^(2/3) + (ax)^(1/3) + a^(2/3)) Do you see that first factor? It's also a factor in the numerator for our limit! To summarize, x^(2/3) - a^(2/3) f'(a) = lim -------------------- x->a x - a (x^(1/3) - a^(1/3)) * (x^(1/3) + a^(1/3)) = lim ------------------------------------------------------ x->a (x^(1/3) - a^(1/3)) * (x^(2/3) + (ax)^(1/3) + a^(2/3)) (x^(1/3) + a^(1/3)) = lim -------------------------------- x->a (x^(2/3) + (ax)^(1/3) + a^(2/3)) Taking the limit should now be straightforward. I hope this helped! Let us know if you have any more questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/