Finding minima and maxima of a specific expression

Question

I found this expression while solving my assignment :

$$ \cos(2x) + \sqrt{{\cos(2x)}^2+ 15}$$

While I found maxima easily by maximising both terms by taking cos(2x) = 1, I am finding it difficult to find minima. I know basic calculus, and differentiating it or plotting will definitely give me an answer, but I feel there is a shorter answer using inequalities. May be Cauchy, but I am unable to apply it, so is there any method to find minima of the expression using inequalities, or quadratics?

Answer 1

If we set $f(x)=\cos(2x) + \sqrt{{\cos(2x)}^2+ 15}$ and use $\cos(y+\pi)=-\cos y$, we have $f(x)\,f(x+\pi/2)=15$, so if we have the maximum (and that's easy, since it's a monotone function of $\cos 2x,$ we also have the minimum.

Answer 2

After making the substitution $y = \cos 2x$, we get $y + \sqrt{y^2+15}$, with bounds $-1 \le y \le 1$.

We guess that this function is increasing, in which case $y = -1$ should be the minimum. So let's write a lower bound that's tight at $y = -1$.

We have, for all $y \in [-1,1]$, \begin{align*} y + \sqrt{y^2+15} &\ge y + \sqrt{y^2 + 15 - 6(y+1)} \\ &= y + \sqrt{y^2 - 6y + 9} \\ &= y + |y - 3| \\ &= y + (3-y) = 3. \end{align*} As a matter of fact, $y + \sqrt{y^2+15} = 3$ when $y = -1$, and we've just confirmed that this is the best possible. So $\cos 2x = -1$ minimizes your original function.

Or we could take derivatives...

Answer 3

Another observation is : $$\cos(2x) + \sqrt{{\cos(2x)}^2+ 15}\\f(x)=x+\sqrt{x^2+15} ,-1\leq x\leq+1$$now find $max\{f(x)\}$ $$f'=1+\frac{2x}{2\sqrt{x^2+15}}=\frac{\sqrt{x^2+15}+x}{\sqrt{x^2+15}}=0\\\sqrt{x^2+15}+x=0\\\sqrt{x^2+15}=-x\\x^2+15=x^2$$there is no root so $$f(-1)=-1+\sqrt{1+15}=3\\f(1)=1+\sqrt{1+15}=5\\3\leq f(x)\leq5 \\min\{f(x)\}=3\\max\{f(x)\}=5$$

Answer 4

Let $\cos2x+\sqrt{\cos^22x+15}=t$.

Thus, $t>0$ and $$t-\cos2x=\sqrt{\cos^22x+15},$$ which after squaring of the both sides gives $$\cos2x=\frac{t^2-15}{2t},$$ which gives $$-1\leq\frac{t^2-15}{2t}\leq1.$$ Now, $$\frac{t^2-15}{2t}\leq1$$ it's $$\frac{t^2-15}{2t}-1\leq0$$ or $$\frac{(t-5)(t+3)}{t}\leq0,$$ which with $t>0$ gives $$0<t\leq5.$$ $$-1\leq\frac{t^2-15}{2t}$$ it's $$0\leq1+\frac{t^2-15}{2t}$$ or $$\frac{(t+5)(t-3)}{t}\geq0,$$ which with $t>0$ gives $$t\geq3.$$ Finally, we obtain $$3\leq t\leq5$$ and we are done!