Find the minimum of function $$f(x)=\frac{4x^2}{\sqrt{x^2-16}}$$ without using the derivative.

In math class we haven't learnt how to solve this kind of problems (optimization) yet. I already know that is solvable using derivatives, but there should be another way. Thanks in advance!

  • $\begingroup$ math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Saucy O'Path Jun 17 '18 at 9:31
  • $\begingroup$ thanks. it's seems too complicate $\endgroup$ – d4ta l0l Jun 17 '18 at 9:34
  • 2
    $\begingroup$ I concur. Nowadays even having others do your stuff for free has become hard. Oh, the humanity! $\endgroup$ – Saucy O'Path Jun 17 '18 at 9:47
  • $\begingroup$ If you can't memorize all this syntax, use: wiris.com/editor/demo/en/developers and copy the mathjax between "$$" to display your question. $\endgroup$ – C. Cristi Jun 17 '18 at 9:51
  • $\begingroup$ thanks Cristi. O'path (empty space) $\endgroup$ – d4ta l0l Jun 17 '18 at 9:59

Since the quantity inside root can't be negative for real values of x we have $$x\in (-\infty, -4]\cup [4,\infty) $$

Hence we substitute $x=4\sec \theta$ for some arbitrary $\theta$

Hence we need to find the minimum value of $$\frac {64\sec^2\theta}{4\tan \theta}=\frac {32}{2\sin\theta\cos\theta}$$

Now using$$2\sin\theta\cos\theta=\sin 2\theta$$

We need to find minimum value of $$\frac {32}{\sin 2\theta}$$ which is simply $32$ attained when $\sin 2\theta=1$ , i.e.$$\theta=\frac {(4n+1)\pi}{4}$$ hence at that point we have $$x=4\sec\theta=\pm 4\sqrt 2$$

  • $\begingroup$ Nice answer! What is the intuition behind the substitution? $\endgroup$ – C. Cristi Jun 17 '18 at 9:44
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    $\begingroup$ @C. Cristi You might know that during integration when we have a form of $\sqrt {x^2-a^2}$ we substitute $x=a\sec\theta$ and as I love integration so much, the thought of substitution was quick enough $\endgroup$ – Rohan Shinde Jun 17 '18 at 9:50
  • $\begingroup$ yes. 32 is the answer, but problem is i can't understand idea of your answer because i'm pupil and i don't know all this things $\endgroup$ – d4ta l0l Jun 17 '18 at 9:55
  • $\begingroup$ @d4ta I0I What exactly do you refer to as "pupil"? Since you have asked a question a referring to inequalities like the AM-GM then it's quite obvious that we might suppose you know some basic trigonometry. Which grade do you study in? $\endgroup$ – Rohan Shinde Jun 17 '18 at 9:58
  • $\begingroup$ i don't study in any grade. i'm study in middle school. $\endgroup$ – d4ta l0l Jun 17 '18 at 10:01

Rewrite $f(x)=\frac {4x^2}{\sqrt{x^2-16}}=4(\sqrt{x^2-16}+\frac{16}{\sqrt{x^2-16}})$

Using the AM-GM inequality we get:


then by multiplying by 2 we get:


But is it attainable? Well we have equality when $\sqrt{x^2-16}=\frac{16}{\sqrt{x^2-16}}\implies x=\pm4\sqrt2$. And yes, it is attainable.

  • $\begingroup$ at first thank you for this answer. this solution is quite unordinary and new to me. can i ask you why you've used AM-GM inequality? $\endgroup$ – d4ta l0l Jun 17 '18 at 10:52
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    $\begingroup$ because I saw that the function can be rewritten such that we will get a "constant product" since GM is $\sqrt{ab}$ if we have $a=1/b$ then we have a constant product. $\endgroup$ – C. Cristi Jun 17 '18 at 10:58

Re write as $k= (4x^2)/(x^2 -16)^.5 $,

$k^2 * ( x^2 -16) = 16x^4$

This is a quadratic in x^2. When the discriminant is 0, only one solution for x^2 is implied.As squaring creates extra solutions, check which of them is implied by your function. This is either a global minimum or maximum


It is the same as finding the minimum of $\frac{4z}{\sqrt{z-16}}$ for $z>16$, or the minimum of $\frac{16 t}{\sqrt{t-1}}$ for $t>1$, or the minimum of $\frac{16(u+1)}{\sqrt{u}}$ for $u>0$, or the minimum of $16\left(v+\frac{1}{v}\right)$ for $v>0$. It is clearly $\color{red}{32}$ by the AM-GM inequality.


Substitution makes it easier: Write $t=x^2-16$ then using inequality between AG and GM we have $$\frac{4x^2}{\sqrt{x^2-16}}=\frac{4t+64}{\sqrt{t}}\geq {4\sqrt{16t}\over \sqrt{t}}=16$$

and the minimum is achieved at $t=16$ or $x =\pm 4\sqrt{2}$


The solution can be arrived at by using the quadratic formula.

To begin, we want to analyze how $f(x)=\frac{4x^2}{\sqrt{x^2-16}}$ on the domain $x \gt 4$ can take on any value $h \gt 0$:

$\tag 1 f(x) = h \text{ iff } 16 x^4 - h^2 x^2 + 16 h^2 = 0$

For this quadratic equation in $x^2$ to have any solutions the discriminant must be greater than or equal to zero:

$\tag 2 h^4 - (4)(16)(16 h^2) \ge 0 \text{ iff } h^2 \ge (4)(16)(16) \text{ iff } h \ge (2)(16) \text{ iff } h \ge 32$.

So the range of the function is the interval $[32, +\infty)$ and so the minimum value hit by $f$ is $32$.

Using $\text{(1)}$ and $\text{(2)}$ above and the quadratic formula when the discriminant is zero, we see that

$\tag 3 x^2 = \text{[negative } b \text{ divided by } 2 a \text{]} = (32^2)/(2)(16) \text{ iff } x = 32/(4 \sqrt 2) = 4 \sqrt 2$ so

$\tag 4 f(4 \sqrt 2) = 32$

See Ethan Horsfall's answer.




Here, $a$ is the minimum function value.

Square both sides and convert the resulting polynomial relation to the form $P(x)=0$:



Now if $a$ is to be the minimum function for $f(x)$ value then the polynomial will have a squared factor for that value of $a$; the value of $x$ where that occurs (called $x_0$) is the resulting double root:




For this factorization to hold The cubic and linear terms must satisfy $p-2x_0=0$ and $-2q+px_0=0$ (the function would not be defined at $x_0=0$). Therefore $p=2x_0, q=x_0^2$. Put these into the quadratic and constant terms:



Dividing the second of these two equations by the first leads to $x_0^2=32$, thus $x_0=\pm 4\sqrt{2}$ for the minimizing $x$ value. The minimum function value, which is clearly positive, is then $a=\sqrt(2x_0^2)=8$.


A bit of trickery:

Note :$|x|>4 .$

Set $y= (x^2-16)^{1/2}$, $y>0$.

With $x^2= y^2 +16$:

$F(y): =\dfrac {4y^2 +64}{y}=$

$4y +64/y =$

$4(√y -4/√y)^2+ (4)(8)$.

$F_{min}(y)= 32$; at $y= 4.$

Corresponding $x$:

$y^2+16 = x^2,$ or

$x= ^+_-(32)^{1/2}$

  • $\begingroup$ I posted this idea an hour ago. $\endgroup$ – C. Cristi Jun 17 '18 at 11:12

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