How many extremum does the function have for given values of a and b?

[Saitama]

Homework Statement

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

Homework Equations

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!

 

[Dick]

Homework Statement

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

Homework Equations

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!

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

 

[Saitama]

If f'(x) has two roots, then what?

f(x) would have two extremum then.

 

[Dick]

f(x) would have two extremum then.

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

 

[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

 

[Dick]

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

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.

 

[ehild]

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.

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

 

[Dick]

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

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.

 

[Saitama]

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?

 

[Dick]

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?

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

 

[Saitama]

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

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?

 

[Dick]

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?

Yes.

 

[ehild]

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?

Add that x²-x+1 is always positive.

ehild

 

[Saitama]

Yes.

Thanks a lot Dick! :)

Add that x²-x+1 is always positive.

ehild

Yes, I did notice that, thanks ehild.