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I have a function $f(x)$: $$f\left(x\right)=\frac14x^\left(2/3\right)-x^\left(1/3\right)$$

For which I have found the First derivative: $$f'\left(x\right)=\frac{1}{6x^\left(1/3\right)}-\frac{1}{3x^\left(2/3\right)}$$

Now, the Critical Numbers I have calculated are: $x=0$ and $x=8$.

However, when I plug them in the first derivative and sketch the number line, I get: $$Local-Maxima=D.N.E.$$ $$Local-Minima=8$$ It appears the Local Minima is wrong, but I don't understand how it is possible. The number line shows me that the function decreases in the interval $(-\infty,0)\cup(0,8)$ and then increases in the interval $(8,\infty)$.

Thanks in advance for the help!

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2 Answers 2

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$x=8$ is the location of a local minima and also the absolute minimum. If $f$ is decreasing (going downwards) from the left, hits a critical point, and then starts increasing (going upwards), then this is justification for that critical point being a local minimum.

More rigorous: $f'(x)=0$ or is undefined at $x=0,8$. Now $f'<0$ on $(-\infty,8)\setminus\{0\}$, so $f$ is decreasing on that interval. $f'(8)=0$ so $x=8$ is the location of a critical point. And $f'>0$ on $(8,\infty)$, so $f$ is increasing on that interval.

Absolute Minimum: We can reason that $f(8)$ is the absolute minimum by comparing $f(0),\,f(8)$ and the limits at infinity.

\begin{align*} L &= \lim_{x\to -\infty} f(x) = \infty \\ R &= \lim_{x\to \infty} f(x) = \infty \\ f(0) &= 0 \\ f(8) &= -1 \end{align*}

So $f(8)<f(0)<L=R$, which implies $f(8)$ is the absolute minimum of $f$.

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  • $\begingroup$ Thank you for your answer! So technically I am right, although WebAssign tells me otherwise? $\endgroup$
    – Lollators
    Jul 28, 2017 at 13:32
  • $\begingroup$ @Lollators Not sure what WebAssign is, but both Mathematica and Desmos agree that the minimum is $f(8)$. $\endgroup$
    – Dando18
    Jul 28, 2017 at 13:36
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we have $$f''(x)=-\frac{1}{18x^{4/3}}+\frac{2}{9x^{5/3}}$$ and $$f''(8)=\frac{1}{288}>0$$ therefore we have an minimum for $x=8$ with the second derivative test.

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