What are the points of inflection, if any, of #f(x)=x^(1/3) #?

For #f(x)=x^{1/3}#, we have #f'(x)=1/3 x^{-2/3}=1/(3x^{2/3})# when #x !=0# and #f''(x)=-2/9 x^{-5/3}=-2/(9x^{5/3})# when #x !=0#. Therefore, #f''(x) > 0# when #x < 0# and #f''(x) < 0# when #x > 0#.

However, #f'(0)# does not exist since #lim_{h->0}(f(0+h)-f(0))/h=lim_{h->0}(h^{1/3})/h=lim_{h->0}1/(h^{2/3})# does not exist. This also implies that #f''(0)# does not exist.

Even though #f'(0)# does not exist, the graph of #f# does have a vertical tangent line at #x=0# and you can see visually that there is an inflection point there as well in the picture below.

To find the inflection points of \( f(x) = x^{1/3} \), we need to calculate its second derivative \( f''(x) \) and then find where it changes sign. First, find the first derivative: \[ f'(x) = \frac{d}{dx}(x^{1/3}) = \frac{1}{3}x^{-2/3} \] Now, find the second derivative: \[ f''(x) = \frac{d}{dx}(\frac{1}{3}x^{-2/3}) = -\frac{2}{9}x^{-5/3} \] The second derivative is negative for all \( x \neq 0 \), indicating that the graph of \( f(x) = x^{1/3} \) is concave down for \( x \neq 0 \). However, since the second derivative does not exist at \( x = 0 \), we need to check the behavior of the function around \( x = 0 \). By examining the behavior of the function, we see that it changes concavity at \( x = 0 \), indicating a possible inflection point. To confirm, we can observe the signs of the second derivative around \( x = 0 \). For \( x > 0 \), \( f''(x) < 0 \), indicating concave down. For \( x < 0 \), \( f''(x) > 0 \), indicating concave up. Therefore, \( x = 0 \) is an inflection point of \( f(x) = x^{1/3} \).