Derivative Applications: Exploring Roots, Concavity, and Inflection Points

[kingwinner]

I am stuck on the following questions lately. I hope someone can explain/help me out, any little help is greatly appreciated!

1) Let f(x) = x^3 - 3x + b.
1a) Show that f(x)=0 for at most one number x in [-1,1].
1b) Determine the values of b such that f(x)=0 for some x in [-1,1].

This is an example in my textbook,
For part a, they showed f'(x)=0 at x=+/-1, so f'(x) not=0 on (-1,1), so f(x) = 0 has at most one root (otherwise, by rolle's theorem,...)
For part b: they said that when f(x)=0, b = 3x - x^3 = x (3 - x^2). When x E [-1,1], |x (3-x^2)|<=2. Thus |b|<=2
"When x E [-1,1], |x (3-x^2)|<=2" <------This is where I can't follow this logic, can someone please explain this?




2) Let f and g be differentiable functions on the interval (0,c) such that f(0)=g(0). Prove that if f'(x)>g'(x) for all x E (0,c), then f(x)>g(x) for all x E (0,c).

My attempt:
Let F(x)=f(x)-g(x), I need to prove that F(x)>0
F'(x) = f'(x)-g'(x) > 0 on x E (0,c) since f'(x)>g'(x)
So F increases on (0,c)
This is not what I want...how can I actually prove that F(x)>0?





3) Determine A and B so that the graph of f(x) = Acos(2x) + Bsin(3x) will have a point of inflection at (pi/6,5).

My solution: (incompleted)
(pi/6,5) must lie on the curve, so I sub. (pi/6,5) into f(x), getting A+2B=10 ...(1)

f''(x) exists for all x E R, if there is a point of inflection at x=pi/6, then it must be true that f''(pi/6)=0, so I set f''(pi/6)=0, obtaining -2A-9B=0 ...(2)

Solve (1) and (2), I get A=18,B=-4

BUT the most important thing about point of inflection is that f''(x) MUST change sign (change concavity),f''(x)=0 does not mean that a certain point is a point of inflection for sure, so now I must show that in a small neighbourhood around x=pi/6, f'' has opposite sign, sub. A=18,B=-4 into f''(x), I get f''(x)=-36[2cos(2x)-sin(3x)], but now how can I prove that the 2nd derivative will change sign at x=pi/6? The trouble is that I don't know the location of the other inflection points, so I can't really use the "test sign" method (because I don't know in what interval will the sign of f'' keep constant), say if I evalutate f''(0) to test sign, the trouble is that between 0 and pi/6, f'' may not be keeping a constant sign...how can I take care of this mess?

 

[cristo]

I'll have a go at 1:

I am stuck on the following questions lately. I hope someone can explain/help me out, any little help is greatly appreciated!

1) Let f(x) = x^3 - 3x + b.
1a) Show that f(x)=0 for at most one number x in [-1,1].
1b) Determine the values of b such that f(x)=0 for some x in [-1,1].

This is an example in my textbook,
For part a, they showed f'(x)=0 at x=+/-1, so f'(x) not=0 on (-1,1), so f(x) = 0 has at most one root (otherwise, by rolle's theorem,...)
For part b: they said that when f(x)=0, b = 3x - x^3 = x (3 - x^2). When x E [-1,1], |x (3-x^2)|<=2. Thus |b|<=2
"When x E [-1,1], |x (3-x^2)|<=2" <------This is where I can't follow this logic, can someone please explain this?

If you look at x(3-x²), considering we are in the range [-1,1], what are the maximum and minimum values of this expression?

 

[cristo]

3) Determine A and B so that the graph of f(x) = Acos(2x) + Bsin(3x) will have a point of inflection at (pi/6,5).

My solution: (incompleted)
(pi/6,5) must lie on the curve, so I sub. (pi/6,5) into f(x), getting A+2B=10 ...(1)

f''(x) exists for all x E R, if there is a point of inflection at x=pi/6, then it must be true that f''(pi/6)=0, so I set f''(pi/6)=0, obtaining -2A-9B=0 ...(2)

Solve (1) and (2), I get A=18,B=-4

BUT the most important thing about point of inflection is that f''(x) MUST change sign (change concavity),f''(x)=0 does not mean that a certain point is a point of inflection for sure, so now I must show that in a small neighbourhood around x=pi/6, f'' has opposite sign, sub. A=18,B=-4 into f''(x), I get f''(x)=-36[2cos(2x)-sin(3x)], but now how can I prove that the 2nd derivative will change sign at x=pi/6? The trouble is that I don't know the location of the other inflection points, so I can't really use the "test sign" method (because I don't know in what interval will the sign of f'' keep constant), say if I evalutate f''(0) to test sign, the trouble is that between 0 and pi/6, f'' may not be keeping a constant sign...how can I take care of this mess?

To show if a critical point is a point of inflection, look at the values of f'(x) either side of the critical point. If the are opposite in sign, then you have a point of inflection.

 

[AlephZero]

2) You also know the value of F(0).

 

[kingwinner]

I'll have a go at 1:



If you look at x(3-x²), considering we are in the range [-1,1], what are the maximum and minimum values of this expression?

How can I calculate this? The maximum and minimum may not occur at the endpoints...

 

[kingwinner]

To show if a critical point is a point of inflection, look at the values of f'(x) either side of the critical point. If the are opposite in sign, then you have a point of inflection.

Yes, I know this fact perfectly, but my problem with question 2 I don't think it's possible to do so without knowing ALL the points where the 2nd derivative may possibly change sign...

 

[kingwinner]

2) You also know the value of F(0).

(i) F'(x) = f'(x)-g'(x) > 0 on x E (0,c)
(ii) F increases on the OPEN interval (0,c)
(iii) F(0)=f(0)-g(0)=0

Do the 3 conditions above guarantee that F(x)>0 for all x E (0,c)? What if F is discontinuous at 0? (such as jump discontinuity), my thought is that this won't mean that F(x)>0 for all x E (0,c), is there any missing information in the actual question itself?

 

[kingwinner]

Can anyone explain a little bit further? Any help is appreciated!