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For example this function: $f(x) = (x + 1)|x + 3| + 2$ it has no derivative in $x = -3$, but how can I discover it?

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Do you mean not differentiable? – andybenji Dec 27 '12 at 20:43
That's possible. I don't know these words in english exactly – user50222 Dec 27 '12 at 20:49
Draw a picture!!! Look for kinks. – copper.hat Dec 27 '12 at 20:55
I'm not asking for recognizing acording to a graph... – user50222 Dec 27 '12 at 20:57
up vote 4 down vote accepted

Recall that the $\vert y \vert$ is not differentiable at $y=0$. Hence, in your case, $f(x) = (x+1) \vert x + 3\vert + 2$ i.e. $$f(x) = \begin{cases} (x+1)(x+3) + 2 & x \geq -3\\ -(x+1)(x+3) + 2 & x \leq -3 \end{cases}$$ Hence, the only point where derivative need not exist is at $x+3 = 0$ i.e. $x=-3$, since away from $x=-3$, the function is a nice quadratic function.

At $x=-3$, we have \begin{align} \lim_{h \to 0^+} \dfrac{f(-3+h) - f(-3)}h & = \lim_{h \to 0^+} \dfrac{((-3+h+1)(-3+h+3)+2) - ((-3+1)(-3+3)+2)}{h}\\ & = \lim_{h \to 0^+} \dfrac{h(h-2) +2 - 2}h = \lim_{h \to 0^+} \dfrac{h(h-2)}h = -2 \end{align} \begin{align} \lim_{h \to 0^-} \dfrac{f(-3+h) - f(-3)}h & = \lim_{h \to 0^-} \dfrac{(-(-3+h+1)(-3+h+3)+2) - (-(-3+1)(-3+3)+2)}{h}\\ & = \lim_{h \to 0^-} \dfrac{h(2-h)+2 - 2}h = \lim_{h \to 0^-} \dfrac{h(2-h)}h = 2 \end{align}

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The graph of $y=|x+3|$ is just the graph of $y=|x|$ shifted $3$ units to the left, so it has a sharp corner at $x=-3$, and the function $g(x)=|x+3|$ therefore cannot have a derivative at $x=-3$. When $x$ is very close to $-3$, $x+1$ is approximately $-2$, and $(x+1)|x+3|$ is approximately $-2|x+3|$, which also has a sharp corner at $x=-3$. Finally, adding $2$ does nothing but move the graph $2$ units vertically, so it doesn’t smooth out any corners.

This reasoning doesn’t prove that $f$ has no derivative at $x=-3$, because it’s just barely conceivable that multiplying $|x+3|$ by $x+1$ could smooth out the sharp corner, but it’s a very good indication that that’s where you should look for a point of non-differentiability. Once you suspect this, you can verify it by more rigorous means.

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Use the fundamental definition of the derivative

$$ f'(-3)=\lim_{x \to -3}\frac{f(x)-f(-3)}{x-(-3)}, $$

and consider the two limits $ \lim_{x\to -3^{+}} $ and $ \lim_{x\to -3^{-}} $ of the expression.

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Rather than calculate the left and right derivatives, one can use a slightly easier-to-work result that I consider in between the solution above and the geometric approach.

Suppose that $f$ is continuous on $\mathbb{R}$ and differentiable except perhaps at $a\in \mathbb{R}$. Suppose further that $\lim_{x\rightarrow a}f'(x)= L<\infty$ exists. Then $f$ is differentiable at $a$ with derivative $f'(a)=\lim_{x\rightarrow a}f'(x)$.

Proof: Let $h>0$. $f$ is continuous on $[a,a+h]$ and differentiable on $(a,a+h)$. Then, by the Mean Value Theorem, $\exists\,c_h\in(a,a+h)$ such that


Now taking the limit as $h\rightarrow 0$ on both sides:

$\lim_{h\rightarrow 0^+}f'(c_h)=\lim_{h\rightarrow 0^+}\frac{f(a+h)-f(a)}{h}$

Now $c_h\rightarrow a^+$ as $h\rightarrow 0^+$ hence the left hand side is equal to $L$:

$\lim_{h\rightarrow 0^+}\frac{f(a+h)-f(a)}{h}=L$

Similarly taking $h<0$ and $h\rightarrow 0^-$ we can show:

$\lim_{h\rightarrow 0^-}\frac{f(a+h)-f(a)}{h}=L$

Hence $f$ is differentiable at $a$ with derivative $f'(a)=L$ $\Box$

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