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
  • 1
    $\begingroup$ @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 $\endgroup$ – Darkrai 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? $\endgroup$ – Darkrai 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
  • 1
    $\begingroup$ 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

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.