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So here, I have the following $$ f(x) = 4^x + 2^x + 1 $$

Now I tried to find the minimum value by doing $y = f(x)$, then doing $4^x + 2^x + 1 - y = 0$, which I solve with the $D \geq 0 $, since this is a real quadratic.

This gives me $y \in [{\frac{3}{4}}, \infty)$, which is not correct here, since $\frac{dy}{dx} = 0$ says me that when $x \rightarrow -\infty$, $y \rightarrow 1$.

Could anyone please explain me why doing with the quadratic way is wrong here, since notice that this is a quadratic of $2^x$?

SomeRandomGuyOnTheInternet
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    There is no minimum value but only a greatest lower bound, which is indeed $\lim_{-\infty}f=1.$ I don't know what you are doing to find $y\ge\frac34.$ Instead, note that $g(t):=t^2+t+1$ is increasing on $(0,+\infty),$ which is the range of the function $g:\Bbb R\ni x\mapsto 2^x.$ This is why $\inf f(\Bbb R)=\inf g((0,+\infty))=g(0).$ – Anne Bauval May 12 '23 at 16:42
  • Oh alright, thank you for your explanation @AnneBauval, I was thinking that this function is always strictly increasing like that in the whole domain of real numbers, which is why there was no minimum point to find. Wanted to clarify it. – SomeRandomGuyOnTheInternet May 12 '23 at 16:55
  • Oops the last function $x\mapsto 2^x$ should not be called also $g.$ It needs no name. + You are right, $f$ is increasing (e.g. as a composition of two increasing functions). – Anne Bauval May 12 '23 at 16:55
  • $t^+t+1$ is a strictly increasing function of $t$ when $t$ is positive. Since you have $t=2^x$, which also a strictly increasing function, as is $x=\log_2(t)$, you will have $t$ being positive for all real values of $x$. – Henry May 12 '23 at 17:10
  • Your idea of replacing $~2^x~$ with a variable, such as $~t~$ (i.e. $~t = 2^x$) is a good one. What you may have overlooked is that when (for example) $~t = 2^x~$ then $~t~$ must always be positive. This is because for all values of $~x~$ (positive, negative, or zero), you have that $~2^x > 0.$ – user2661923 May 12 '23 at 18:58

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Your idea of thinking of $y$ as $t^2+t+1$ is often a really useful technique.

In this case we can write $t^2+t+1=(t+\frac{1}{2})^2+\frac{3}{4}$ which you then correctly state has a minimum of $\frac{3}{4}$.

However, this minimum is attained for $t=-\frac{1}{2}$ and this is not, of course, a possible value for $t=e^x$.

I hope this clarifies why your two methods give apparently 'different' answers.

user1172706
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