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I have that $f(x) = |x^2 -4x|$

What I've done is trying to define $f(x)$ with the zero values being 0 and 4. But not really sure if that's how I'm supposed to go about solving this.

JohnDoe
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2 Answers2

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we have $$f(x)=|x^2-4x|=|x|\cdot |x-4|$$ and if $$x\geq 4$$ then $$f(x)=x(x-4)$$ if $$0\le x<4$$ then we have $$f(x)=x(4-x)$$ if $$x<0$$ we get $$f(x)=-x(4-x)$$ now can make a Image of your function.

Dr. Sonnhard Graubner
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Note that $$\lim_{x\to 0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^+}\frac{|x^2-4x|}{x}=\lim_{x\to 0^+}\frac{-(x^2-4x)}{x}=-\lim_{x\to 0^+}(x-4)=4$$ and $$\lim_{x\to 0^-}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^-}\frac{|x^2-4x|}{x}=\lim_{x\to 0^-}\frac{(x^2-4x)}{x}=\lim_{x\to 0^+}(x-4)=-4$$ What may we conclude about the differentiability of $f$ at $0$? What about the other point $x=4$?

P.S. Recall that $|g(x)|=g(x)$ if $g(x)\geq 0$ and $|g(x)|=-g(x)$ otherwise.

Robert Z
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  • So it's not differentiable at 0 because we get $4 != -4$ but at 4 I get $0 = 0$ so it's differentiable at 4? – JohnDoe Sep 17 '17 at 11:52
  • @JohnDoe For $x=0$ you are correct. As regards $4$ you have to consider $ \lim_{x\to 4^+}\frac{f(x)-f(4)}{x-4}$ and $ \lim_{x\to 4^-}\frac{f(x)-f(4)}{x-4}$. Try again! – Robert Z Sep 17 '17 at 11:57
  • Oh right so then I get $4 != -4$ as well. So does that tell me that $f(x)$ isn't differentiable anywhere? – JohnDoe Sep 17 '17 at 12:01
  • No, it is not differentiable just at $0$ and $4$, because to obtain different limits from the left and from the right we need a change of the sign of $x(x-4)$. – Robert Z Sep 17 '17 at 12:03
  • Ok everywhere except at 0 and 4 then? Isn't the ${x\to 0^+}$ supposed to be $(x^2 - 4x)$ and $x\to 0^-$ the $-(x^2 -4x)$ ? Or am I looking at the graph wrong – JohnDoe Sep 17 '17 at 12:28
  • $(x^2−4x) =x(x-4)$ is negative in $(0,4)$. Hence $|x^2−4x|=-(x^2-4x)$ in a right neighborhood of $0$. – Robert Z Sep 17 '17 at 12:40