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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?

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Because $f''(1)<0$, this point is a local maximum, not an inflection point, and the book is wrong.

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$x=1$ is not an inflection point (as you can see from the graph below).

enter image description here

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).

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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.

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