1
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

$x=8$ 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 $f$.

$\endgroup$
  • $\begingroup$ Thank you for your answer! So technically I am right, although WebAssign tells me otherwise? $\endgroup$ – Lollators Jul 28 '17 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 '17 at 13:36
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.