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I have a hard time solving these kinds of problems, here is an example. For which values of a and b is the following function differentiable at all points? $$f(x)=\sin(|x^2+ax+b|)$$ Thanks in advance.

zz20s
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Xelak
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Whenever the quadratic isn't positive, namely when $\;\Delta =a^2-4b\le0\;$ . Can you see what's the problem with the absolute value of something that vanishes?

Added: suppose the quadratic vanishes at $\;x_0< x_1\;$ , so that $\;x^2+ax+b=(x-x_0)(x-x_1)\;$ .

Observe that if $\;h<0\;$ is such that $\;x_0-x_1+h>0\;$ , then $\;f(x_0+h)=\sin|h(x_0+h-x_1)|\;$ , with $\;h(x_0+h-x_1)<0\;$ .

Now, by definition:

$$\lim_{h\to 0^-}\frac{f(x_0+h)-f(x_0)}h=\lim_{h\to0^-}\frac{sin(|h(x_0+h-x_1)|)-\overbrace{0}^{=f(x_0)}}{h}=$$

$$=\lim_{h\to 0^-}\,(x_0+h-x_1)\frac{\sin\left(-h(x_0+h-x_1)\right)}{h(x_0+h-x_1)}=\color{red}-\lim_{h\to 0^-}\,(x_0+h-x_1)\frac{\sin\left(h(x_0+h-x_1)\right)}{h(x_0+h-x_1)}=$$

$$=x_1-x_0$$

whereas

$$\lim_{h\to 0^+}\frac{f(x_0+h)-f(x_0)}h=\lim_{h\to0^+}\frac{\sin(|h(x_0+h-x_1)|)-\overbrace{0}^{=f(x_0)}}{h}=$$

$$=\lim_{h\to 0^+}\,(x_0+h-x_1)\frac{\sin\left(h(x_0+h-x_1)\right)}{h(x_0+h-x_1)}=x_0-x_1$$

Thus, both one sided limits aren't equal so the limit definint the derivative at $\;x_0\;$ doesn't exist and thus $\;f'(x_0)\;$ doesn't exist.

Now you try to do something similar when $\;x_0=x_1\;$...but this time you'll find out the function is differentiable at $\;x_0\;$ !

DonAntonio
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  • Thanks for the answer, however I don't quite get it, could you explain in more detail? – Xelak Apr 23 '16 at 10:47
  • @Kalex Read now the added part and try to fill in details. – DonAntonio Apr 23 '16 at 11:02
  • Double root...cough, cough. – JP McCarthy Apr 23 '16 at 12:15
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    @JpMcCarthy Thank you and drink some water. What about the double root case? – DonAntonio Apr 23 '16 at 12:18
  • @Joanpemo isn't the function differentiable at all points when the quadratic has a double root at the origin? – Xelak Apr 23 '16 at 12:37
  • @Kalex Yes, it is...and I addreses this at the end of my answer above, and as far as I can see now of the top of my head it doesn't matter where the quadratic has the double root. – DonAntonio Apr 23 '16 at 13:00
  • @Joanpemo shouldn't the answer then be $a^2-4b≥0;$? – Xelak Apr 23 '16 at 13:06
  • @Kalex No, because $;\Delta>0\implies;$ there are two different (real) roots, and that means the function isn't differentiable at these points . When there's a unique root, or when there are no roots at al, i.e. when $;\Delta\le0;$, we have differentiability i nthe whole real line. – DonAntonio Apr 23 '16 at 13:14