Analysis and the DerivativeDate: 01/23/2001 at 14:19:58 From: Yvonne Subject: Analysis and the Derivative In my analysis class, we have come to the chapter on the derivative. The text suggests considering the following problem. I'm having a little trouble starting and I would greatly appreciate it if you could help me a little. Suppose that f:RtoR is differentiable at c and that f(c)=0. Show that g(x):=|f(x)| is differentiable at c if and only if f'(c)=0. Any suggestions would be terrific. Thanks. Date: 01/24/2001 at 21:36:14 From: Doctor Fenton Subject: Re: Analysis and the Derivative Dear Yvonne, Thanks for writing to Dr. Math. For this problem, you need to use the definition of derivative: a function F(x) is differentiable at x = a if F(x)-F(a) lim --------- exists. The value of this limit is F'(a). x->a x - a So, you know that (1) f(c)=0 ; and (2) f is differentiable at x = c, or in other words, f(x)-f(c) lim --------- = f'(c). x->c x - c I think you'll need the epsilon-delta version of this: (I'll write e for epsilon and d for delta) Given e>0, there is a d>0 such that for all x such that 0<|x-c|<d, | f(x)-f(c) | | ---------- - f'(c)| < e . | x - c | To show g(x) is differentiable at x = c, you must show that g(x)-g(c) lim --------- exists. x->c x - c That is, you must find a number L such that, given any e > 0, you can find a d>0 such that if x satisfies 0 < |x-c| < d, then | g(x)-g(c) | | ---------- - L | < e . | x - c | If you look at a function near a point where it crosses the axis, and if the function has a non-zero slope there (so f'(c) =/= 0, where =/= means "is not equal to"), then the graph will look like \ / / \ / ------/----- , so g(x) = |f(x)| looks like ---------- , / because taking the absolute value reflects or folds the part of the graph below the x-axis up into the part above the axis. That intuitively explains why f'(c) must = 0 in order for |f(x)| to be differentiable at c. You should be able to see that, in that case, the derivative of g at c, g'(c), should also be 0, so we expect L=0. Since f(c)=0, and if we assume that f'(c)=0, then we know that (putting in 0 for f(c) and f'(c) in the e-d definition above, Given e>0, there is a d>0 such that for all x such that 0<|x-c|<d, | f(x) | | ---------- | < e . | x - c | We must consider | g(x)-g(c) | |---------- - (0) | | x - c | , since as I indicated, we expect g'(c)=0 . But g(c)=|f(c)| = 0 , so this is | |f(x)| | | f(x) | |-------- | = |-------| | x - c | | x - c | , but this is exactly the expression above which we know is < e. So the d you chose for f also works for g. For the converse, it's easier to use contrapositive reasoning. Assume that f'(c) =/= 0 . Then the two-sided limit f(x)-f(c) lim --------- = f'(c) exists, x->c x - c so in particular, the two one-sided limits f(x)-f(c) f(x)-f(c) lim --------- = f'(c) , and lim --------- = f'(c) , x->c+ x - c x->c- x - c or f(x) f(x) lim --------- = f'(c) , and lim --------- = f'(c) , x->c+ x - c x->c- x - c since f(c) = 0. But then (remember that g(c)=0) g(x) |f(x)| lim --------- = lim --------- , x->c+ x - c x->c+ x - c and g(x) |f(x)| lim --------- = lim --------- . x->c- x - c x->c- x - c But if f'(c) =/= 0, then it is either positive or negative. Assume it is positive. Then f(x)-f(c) lim --------- = f'(c) x->c+ x - c > 0 , so all the quotients f(x)/(x-c) must be positive, for x close to c. The denominator is clearly positive, so the numerator must also be positive, and f(x)>0 for x>c (i.e. in some small interval c<x<c+d). Similarly, f(x)-f(c) lim --------- = f'(c) x->c- x - c > 0 , and now the denominators are all negative, so the numerator must also be negative, for x to the left of c. But g(x)=|f(x)| is either f(x) (if f(x) is positive), or g(x) is -f(x) (if f(x) is negative). g(x) |f(x)| f(x) lim --------- = lim --------- = lim ---------- = f'(c), x->c+ x - c x->c+ x - c x->c+ x - c and g(x) |f(x)| - f(x) lim --------- = lim --------- = lim --------- = - f'(c). x->c- x - c x->c- x - c x->c- x - c If f'(c) =/= 0, the two one-sided limits are unequal, so the two-sided limit does not exist. If you have further questions, please write again. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ |
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