Prove that f(x)>=0 for all x in R iff b^2-ac=<0

[Robb]

Homework Statement

Homework Equations

The Attempt at a Solution

This is where I'm at and I'm not sure what my next step is. Any help would be much appreciated!

 

[member 587159]

Homework Statement

View attachment 237693

Homework Equations

The Attempt at a Solution

This is where I'm at and I'm not sure what my next step is. Any help would be much appreciated!
View attachment 237695

Great work so far! You made a sign mistake at the last equality though.

You want that $f(x) \geq 0$ for all $x \in \mathbb{R}$, or equivalently (why??) $\min f = \frac{-b^2 +ac}{a} \geq 0$. Can you proceed? Note that the hypothesis $a>0$ is essential to proceed.

 

[Robb]

This is where I'm not sure. I tried setting min f = (-b^2+ac)/a and solving for x but I'm not sure if that tells me anything and it doesn't relate to the hypothesis. I guess I'm unsure of how to use the hypothesis.

 

[member 587159]

This is where I'm not sure. I tried setting min f = (-b^2+ac)/a and solving for x but I'm not sure if that tells me anything and it doesn't relate to the hypothesis. I guess I'm unsure of how to use the hypothesis.

That expression doesn't depend on $x$...

It really is one line to get the right answer from here.

 

[Robb]

Does it have to do with divisibility of a, which must be greater that zero?

 

[member 587159]

Does it have to do with divisibility of a, which must be greater that zero?

Yes, certainly.

When is the following statement true?

$ax \geq 0 \iff x \geq 0$

 

[Robb]

for all x in R?

 

[member 587159]

for all x in R?

I forgot to mention that $x \in \mathbb{R}$ is fixed. I want a condition for $a$.

 

[Robb]

(a) must be greater than zero but that's a given

 

[member 587159]

(a) must be greater than zero but that's a given

Yes, indeed. Apply this and you will be done (in fact, you have to apply it to $1/a$ instead of $a$)

 

[Robb]

How does b^2-ac =< 0 tell us that f(x) >= 0?

 

[member 587159]

How does b^2-ac =< 0 tell us that f(x) >= 0?

$b^2 - ac \leq 0 \implies \frac{ac -b^2}{a} \geq 0$

Since $\frac{ac -b^2}{a} = \min f$, we know that $\min f \geq 0$. It is now obvious that $f \geq 0$, as $f \geq \min f \geq 0$

 

[Robb]

NOW I GET IT! I was looking at min f as f' min. I assume I could also apply the second derivative test for extrema? So, f" = 2a, and we know a>0, hence f is positive?