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The function $f:\Bbb R\to\Bbb R$ defined by $f(x)=\min(3x^3+x,|x|)$ is

(A) continuous on $\Bbb R$, but not differentiable at $x=0$.
(B) differentiable on $\Bbb R$, but $f\,'$ is discontinuous at $x=0$.
(C) differentiable on $\Bbb R$, and $f\,'$ is continuous on $\Bbb R$.
(D) differentiable to any order on $\Bbb R$.

My attempt: Here,$f(x)=|x|,x>0$;$f(x)=3x^3+x,x<0$;$f(x)=0$ at $x=0.$ Also,$Lf'(0)=Rf'(0)=1.$ So,$f$ is differentiable at $x=0.$ But I am having trouble to check whether $f'$ is continuous at $x=0$ or not. Can someone point me in the right direction?Thanks in advance for your time.

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If you check $x \ge 0$ you see that there $\lvert x \rvert \le 3 x^3 + x$, and for $x < 0$ you have $\lvert x \rvert > 3 x^3 + x$, so your function is $3 x^3 + x$ for $x < 0$ and $x$ for $x \ge 0$. It is continuous at $x = 0$ (both branches agree on that point, essentially). For derivatives, similarly check both branches and see if they agree. – vonbrand Feb 9 '13 at 16:51

2 Answers 2

up vote 1 down vote accepted

We have $$ \lim_{x\rightarrow 0^+}\frac{f(x)-f(0)}{x}=\lim_{x\rightarrow 0^+}\frac{|x|}{x}=1 $$ and $$ \lim_{x\rightarrow 0^-}\frac{f(x)-f(0)}{x}=\lim_{x\rightarrow 0^-}\frac{3x^3+x}{x}=1. $$ So $f$ is differentiable at $0$, and moreover $f'$ is continuous at $0$ since $\lim_{0^+}f'(x)=\lim_{0^+}1=1$ and $\lim_{0^-}f'(x)=\lim_{0^-}9x^2+1=1$.

This excludes (A) and (B). Then note that (C) is true since the derivative is $9x^2+1$ for $x>0$ and $1$ for $x\geq 0$.

Now $f$ is obviously differentiable three times on $\mathbb{R}^*$ with $f^{(3)}(x)=18$ for $x<0$ and $f^{(3)}(x)=0$ for $x>0$. So $f^{(3)}$ can't be continuous at $0$. So $f^{(4)}(0)$ does not exists. This excludes (D).

So (C) is the only correct choice.

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I do not understand the line "Then note that (C) is true since the derivative is $3x^2+x$ for $x>0$ and $1$ for $x\geq 0$." I think the derivative is $9x^2+1$ for $x<0$ and $1$ for $x>0.$ Am I right or am i missing something? – user52976 Feb 9 '13 at 17:08
You're absolutely right, sorry. – 1015 Feb 9 '13 at 17:21
@user33640 Ok, I've edited. This should be ok now. – 1015 Feb 9 '13 at 20:04
Thanks julien... – user52976 Feb 10 '13 at 4:20

HINT: If $x<0$, $f\,'(x)=9x^2+1$, so $$\lim_{x\to 0^-}f\,'(x)=\lim_{x\to 0^-}\left(9x^2+1\right)=1\;.$$

Is this the same as $\lim_{x\to 0^+}f\,'(x)$? Are both one-sided limits equal to $f\,'(0)$?

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For, $x>0,f'(x)=1$ and so $\lim_{x\to 0^+}f\,'(x)=1=f'(0).$Hence,$f'$ is continuous at $x=0.$So ,option $(C)$ looks correct.Am I right sir? – user52976 Feb 9 '13 at 16:55
@user33640: You are indeed. – Brian M. Scott Feb 9 '13 at 19:16

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