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The function ${f(x) = {4\over {x^2}}+ x}$ is defined on the domain ${x > 0, x \in \mathbb{R}}$, the set of real numbers.

Find the maximum and minimum values of ${f(x)}$ on the closed interval ${1 \le x \le 4}$.

So my understanding is that stationary points occur when ${f'(x) = 0}$

In a closed interval, the maximum and minimum values of a function are either a stationary point or at an end of an interval.

${f'(x) = {-8x^{-3}} + 1}$ for stationary points

If I was drawing a sketch of this function, I would factorise to find values for x that I would then plug the values into the original function to find y.

I would then draw a nature table to get the feel of the curve.

My questions are:

  1. How do I Factorise ${-8x^{-3} + 1}$ to find values for x
  2. How do I find the maximum and minimum values that incorporates the closed interval?
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    $\begingroup$ $f'(x)$ is not $-8x^{-3}$, it is $-8x^{-3}+1$. That $+1$ matters! $\endgroup$ Commented Dec 28, 2015 at 21:33
  • $\begingroup$ And note that $-8x^{-3} + 1 = 1^3 -(\frac2x)^3$ $\endgroup$
    – MPW
    Commented Dec 28, 2015 at 21:34
  • $\begingroup$ @RoryDaulton thanks, I have updated the question $\endgroup$
    – dagda1
    Commented Dec 28, 2015 at 21:41

1 Answer 1

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So for first question we have that

$$ -\frac{8}{x^3}+1 = 0 $$ $$ -\frac{8}{x^3} = -1 $$ $$ x^3 = 8 $$ $$ x = 2 $$

now plug this into you function and then check the endpoints of the interval, that is $x = 1$ and $x=4$.

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  • $\begingroup$ If I plug 2 into the function then I get a value of 3 or a stationary point at (2, 3). If I then plug 1 and 4 from the closed intervals into the function I get ${ f(1) = 5}$ and ${f(4) = 1{1\over 4}}$. I am confused what to do next with these results, I cannot find anything on google that describes this. Do you know of any links? $\endgroup$
    – dagda1
    Commented Dec 28, 2015 at 22:00
  • $\begingroup$ @dagda1 Maxima and minima are either at the extreme points of the interval or are stationary points in the interior. Note that $f(4)=17/4$ and $3<17/4<5$. $\endgroup$
    – egreg
    Commented Dec 28, 2015 at 22:07
  • $\begingroup$ @egregso f(1) = 5 is the global maxima and f(2) = 3 is the minima? $\endgroup$
    – dagda1
    Commented Dec 29, 2015 at 11:12

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