Minimum value of $2^{2x}+6\cdot 2^x +18$

Question

Minimum value of $$2^{2x}+6\cdot 2^x +18.$$

It looks like the quadratic equation $$t^2+6t+18=(t+3)^2+9.$$

So the minimum is $9$ if we complete the square, but there is no such $x$ which will give us $2^x=-3$.

So the answer should be $18$ as $2^x>0$ for all $x$! Am I correct?

Answer 1

Write:

$$2^{2x}+6\cdot 2^x +18=(2^x+3)^2+9$$

And see that the minimum happens when $2^x \rightarrow 0$ once $2^x >0$.

So in this case we have infimum (not minimum) and it is $18$.

Answer 2

The function $$f(x)=2^{2x}+6\cdot 2^x +18$$ is strictly increasing on $\mathbb{R}$: if $x_1>x_2$ then $$f(x_1)-f(x_2)=(4^{x_1}-4^{x_2})+6\cdot(2^{x_1}-2^{x_2})>0.$$ Therefore the greatest lower bound is $$\inf_{x\in\mathbb{R}}f(x)=\lim_{x\to -\infty}f(x)=18.$$ Note that $f$ does not attain the value $18$ so you can not say that $18$ is the minimum value.

Answer 3

Yes you are. However the value is not a minimum but an infimum because this value is taken after $$\lim_{x\to-\infty}2^{2x}+6\cdot 2^x +18=18$$ (for all negative real $x$ you have $2^{2x}+6\cdot 2^x +18\gt18$)