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Determine the relative extrema and inflection points of $f(x)=|1+x^{\frac{1}{3}}|$.

After breaking the modulus function we get a piecewise function such that $f(x)=1+x^{\frac{1}{3}}$ when $x>-1$ and $f(x)=-1-x^{\frac{1}{3}}$ when $x<-1$. By inspection,we can see $f(-1)<f(-1-h)$ and $f(-1)<f(-1+h)$ implying $-1$ gives local minima at $x=-1$. But derivative of first piecewise function is $\frac{x^{\frac{-2}{3}}}{3}$ and that of the second piecewise function is $-\frac{x^{\frac{-2}{3}}}{3}$. When set both equal to zero, we get $+\infty$ for the first piecewise function and $-\infty$ for the second piecewise function. Now I am confused about these answers. Also, when I get to calculate the second derivative to calculate inflection points,again I am getting $\pm \infty$. So, it will be very helpful if someone provides a solution to it.

N. F. Taussig
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a_i_r
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  • Notice that $f(x) = |1 + x^{1/3}| \geq 0$, with equality holding if and only if $x = 0$. The derivative is never equal to zero. It is defined in the intervals $(-\infty, -1)$ and $(-1, \infty)$. You should be examining the signs of the first and second derivatives in these intervals. Note that the second derivative changes sign in the interval $(-1, \infty)$. – N. F. Taussig Mar 23 '24 at 12:17

1 Answers1

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Observe that in the complex domain in polar representation with $z=x+iy=r e^{i \phi}$

$$z^{1/3} = r^{\frac{1}{3}} \ \left(\cos \left(\frac{\phi}{3}\right) + i \sin \left(\frac{\phi}{3}\right)\right)$$ for $$\phi\to \pm \pi : \quad r^{\frac{1}{3}} \ \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right) \right) =(x^2+y^2)^{\frac{1}{6}} \ \frac{1}{2}\left( 1 \pm i \ \sqrt(3)\right) $$ is not equal to the $-r^{1/3}$ on the negative real line.

Even in the real domain, by

$$ x^{1/3} = \text{sign}(x)\ \left|x\right|^{\frac{1}{3}} $$

the seemingly smoothness is not quite true: $$ \left(x^{1/3}\right)' = 2\ \delta(x)\ \ \left|x\right|^{\frac{1}{3}} + \frac{1}{3}\left|x\right|^{-\frac{2}{3}} * \text{sign}(x)$$

It follows, that working algebraically with powers with real exponents requires clear definitions of domains and working with expressions that allow representation by convergent complex series - here the everywhere convergent exponential series - as the fundamental definition of the generalizion of integer exponents to rationals and reals (leaving out complex exponents completely).

Roland F
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