Maximum and Minimum value of function -8x^2 -3 at interval (-inf, +inf)

Question

Updates: Correcting a mistake, adding MathJax and removing screenshot...

I'm trying to find maximum & minimum value of function -8x^2-3 at an interval (-inf, +inf), which is an open interval by nature.
Taking limits at -inf and +inf results in -inf, i.e. this function has a Maximum or maximum value, but no minimum value.
Finding critical points gives x=0
Evaluating function at x=16 gives -3.
Although, this is the only value, but -ve sign shows, it should be a Minima, not Maxima.

Here is my effort.

Taking Limits: $$\lim{x\to -\infty}=-\infty $$ $$\lim{x\to +\infty}=-\infty $$

Shows that it has a maximum, but no minimum value;

Finding Critical Point: $$ \bar f(x)=-16x $$ Equating to 0: $$ -16x=0 => x=0 $$ Evaluating function at x=0: $$ f(0)=-3 $$

As this contains -ve sign, I think, it is the minimum value. So what I am missing or confusing here?

Answer 1

$$-8x^2-3\leq-3.$$ The equality occurs for $x=0$, which says that $-3$ is a maximal value.

The minimal value does not exist.

Answer 2

Shows that it has a maximum, but no minimum value;

Yes, yes, it does. Put a pin in that: It has a maximum. It has no minimum.

As this contains -ve sign, I think, it is the minimum value.

Well, okay but let's go back to the pin. "It has a maximum. It has no minimum."

So that can't be right.

So what I am missing or confusing here?

Well, Why do think it is a minimum value? You say that because $f(0) = -3$ and $-3$ has a negative sign it must be a minimum.

Okay, why do you think that?

If something is negative, it must be a minimum? It can't be a maximum?

Really?

Can't a negative number ever be a maximum? Can't a function always be negative? After all a if a minimum can be positive for a function that is always positive (such as $g(x) = 8x^2 + 3\ge 3$), couldn't a function that is always negative (such as $f(x) = -8x^2 - 3 \le -3$) have a maximum that is negative?

After all. $x^2 \ge 0$ so $-x^2 \le 0$ so $-8x^2 \le 0$ and $-8x^2 -3 \le -3$. So $-8x^2 - 3$ is always less than $0$ so is there any reason it shouldn't have a maximum that is less than $0$?

So... You did all the work. And you did it all correctly.

$-8x^2 - 3$ has a maximum when $x = 0$ and that maximum is $-8*0^2 - 3 = -3$. And, yes, that maximum is negative. Which is not a contradiction.