# Finding number of extremum

1. Sep 18, 2013

### Pranav-Arora

1. The problem statement, all variables and given/known data
For all $a,b \, \in \, R$, the function $f(x)=3x^4-4x^3+6x^2+ax+b$ has:
a) no extremum
b) exactly one extremum
c) exactly two extremum
d) three extremum

2. Relevant equations

3. The attempt at a solution
$f'(x)=12x^3-12x^2+12x+a=12x(x^2-x+1)+a$
If a=0, there is one extremum but how what about the other values of a? The given answer is b and using a=0 gives the answer. What if for some other value of a there are more than one extremum.

Any help is appreciated. Thanks!

2. Sep 18, 2013

### Dick

Are you sure you've really given this enough thought? If f'(x) has two roots, then what????

3. Sep 19, 2013

### Pranav-Arora

f(x) would have two extremum then.

4. Sep 19, 2013

### Dick

That's pretty obvious, but it's not what I'm looking for. What does it tell you about f''(x)?

5. Sep 19, 2013

### ehild

How the function behaves if x tends to + or - infinity? What is it at x=0? Is it possible that the function has zero or two extrema?

ehild

6. Sep 19, 2013

### Dick

It might have three just from that consideration. But it doesn't. Because it doesn't even have two. That's what I'm trying to nail down first.

7. Sep 19, 2013

### ehild

I know. It cannot have two extrema, as it would go between infinity and minus infinity then. So it can have one or three extrema. In case of more extrema your hint about f" comes in.

ehild

8. Sep 19, 2013

### Dick

Hmm. When I say two extrema, I mean at LEAST two. Not exactly two. That's where the f''(x) comes in. Your hint is appropos to showing there is at least one and ruling out even numbers.

9. Sep 19, 2013

### Pranav-Arora

I really don't understand what's going on here.

$f''(x)=12(3x^2-2x+1)$
I see that f''(x) is always positive so f'(x) is always increasing. But how do I show that f'(x) would have a root?

10. Sep 19, 2013

### Dick

You might want to think about it some more. Or just cheat and think about ehild's message.

11. Sep 19, 2013

### Pranav-Arora

I had $f'(x)=12x(x^2-x+1)+a$, it tends to $-\infty$ as x tends to $-\infty$ and to $\infty$ as x tends to $\infty$. Since f'(x) always increases, it must have one root so exactly one extremum. Is this correct?

12. Sep 19, 2013

### Dick

Yes.

13. Sep 19, 2013

### ehild

Add that x2-x+1 is always positive.

ehild

14. Sep 19, 2013

### Pranav-Arora

Thanks a lot Dick! :)

Yes, I did notice that, thanks ehild.