# Inflection point for function with fractional exponents

Show that $$f(x) = 4x^{1/3}-x^{4/3}$$ has an inflection point at $$x=1$$.

I correctly get $$f'(x) = \frac{4(1-x)}{3x^{2/3}}\implies f''(x)=-\frac{4(x+2)}{9x^{5/3}}$$

It is clear to me that there is an inflection point at $$x=-2$$ since this value of $$x$$ makes the second derivative zero. The text shows that there is also an inflection point at $$x=1$$. I see that this value makes the first derivative $$=0$$, but I don't understand why this causes an inflection point. Can anyone help clarify this point?

Because $$f''(1)<0$$, this point is a local maximum, not an inflection point, and the book is wrong.
$$x=1$$ is not an inflection point (as you can see from the graph below). On the other hand you missed another inflection point $$x=0$$! Perhaps your book mistakenly wrote $$x=1$$ instead of $$x=0$$. The reason you missed $$0$$ is because the function is not differentiable at $$0$$ (and for such points, the condition of $$f''(x)=0$$ for inflection points obviously fails).
There is only one stationary point. Since $$f(0)=0,f(1)=3,f(4)=0$$, $$x=1$$ cannot be an inflection point as $$f$$ increases then decreases.