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You are not logged in. #1 20120625 12:04:54
g(x) = x^(2/3); g'(0)=?The problem states: if g(x)=x^{2/3}, show that g'(0) does not exist. Now lets find left and right limits separately: From this we can see that left and right limits at point 0 are equal by an absolute value, but have opposite signs; therefore, the does not exists. But according to the rules of differentiation or in this case: My problem is: how correlate with ? They are both supposed to be equal to the slope of tangent line, aren't they? What am I missing? Last edited by White_Owl (20120625 12:05:39) #2 20120625 12:35:48
Re: g(x) = x^(2/3); g'(0)=?I don't know if substituting the value of directly into the limit is the right thingto do, but either way, your calculating of the limit is not correct in two ways. First, you reduced it incorrectly to h^(1/3) instead to h^(1/3). Second you claimed that the left and right limits of what you got aren't the same because they have the same absolute value, but different signs, whereas both limits have the same signed value 0 (which doesn't really have a sign, except when used in limits when denoting left and right limits). The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #3 20120625 16:51:00
Re: g(x) = x^(2/3); g'(0)=?hi White_Owl You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #4 20120625 22:27:58
Re: g(x) = x^(2/3); g'(0)=?Hi Bob The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #5 20120625 22:57:14
Re: g(x) = x^(2/3); g'(0)=?hi Stefy, You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #6 20120625 23:14:39
Re: g(x) = x^(2/3); g'(0)=?Still, the f'(0) exists at that point. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #7 20120627 04:43:22
Re: g(x) = x^(2/3); g'(0)=?Ok. And Example #2 does exactly that. Question #2. Yes, I did a mistake in the final step of calculating the limits. Sorry about that. It should be: But still, I have vs . Where did 2/3 disappeared? Question #3. Actually, I do not see why did you say that domain of the function is just positive numbers? I believe the domain is full R. After all, we can do : Not all graphing calculators are smart enough for such function. I am playing with QtiPlot and for "x^(2/3)" it draw the similar image as bob bundy showed. But if I rewrite formula into "(x^2)^(1/3)", I have a more correct graph. Which still has a strange horizontal part near the zero, which does not disappear when I zoom in. Last edited by White_Owl (20120627 04:47:05) #8 20120627 04:48:18
Re: g(x) = x^(2/3); g'(0)=?
Did you mean 'vertical' ? You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #9 20121123 16:10:48
Re: g(x) = x^(2/3); g'(0)=?You are supposed to put (x+h)^(2/3) not (0+h)^(2/3) when finding the derivative using the definitions (f(x+h)f(x))/h. 