Second Derivative of a Function With Absolute Values
Date: 02/27/98 at 17:37:58 From: David Smith Subject: derivative Dr. Math, Let f(x) = |x|^3. So, f(x) is the absolute value of x to the power 3. Find f"(0). The anwer is f"(0) = 0. At first, I tried f(x) = (|x|)(|x|^2). Then, f'(x) = (|x|)'(|x|^2) + (|x|)(|x|^2)' But we know that f(x) is not differentiable at 0. Also, I tried the power rule: f'(x) = 3(|x|)^2*(|x|)' I got stuck at the same point too. How can I get f"(0) = 0? Thanks
Date: 02/28/98 at 21:51:57 From: Doctor Sam Subject: Re: derivative David, I usually begin problems with absolute value by rewriting the function in terms of its pieces. When x >= 0, |x| is just x. When x < 0, |x| is its opposite, or -x. So / x^3 when x >= 0 f(x) = |x|^3 = | \ (-x)^3 when x < 0 Now to find the derivative: / 3x^2 when x > 0 f'(x) = | ? when x = 0 \ -3x^2 when x < 0 The only question is, "What happens at x = 0?" Does f(x) have a derivative there? The limit of (3x^2) as x approaches zero "from above" is 0. Likewise, the limit of (-3x^2) as x approaches zero from below is 0. So this function has a derivative at x = 0, and it is 0. So we can rewrite the derivative: / 3x^2 when x >= 0 f'(x) = | \ -3x^2 when x < 0 Now do the same thing to find the second derivative. / 6x when x > 0 f"(x) = | ? when x = 0 \ -6x when x < 0 As x approaches zero, both 6x and -6x approach zero, so f"(0) = 0. -Doctor Sam, The Math Forum http://mathforum.org/dr.math/
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