F:R->R, the fourth derivative of f is continuous for all x

[sdff22]

F:R-->R, the fourth derivative of f is continuous for all x...

Homework Statement

Suppose f is a mapping from R to R and that the fourth derivative of f is continuous for every real number. If x is a local maximum of f and f"(x)=0 (the second derivative is zero at x), what must be true of the third derivative at x? Fully justify your answer.

The Attempt at a Solution

my thinking so far is that since the forth derivative is continuous for every real number, it should exists for every real number. If the forth derivative exists, it means that the third derivative is continuous for every real number (since we know that if derivative of f exists at a point x, then f is continuous at that point). so all I know is that the third derivative is continuous at every point including point x.

 

[HallsofIvy]

You should know more than that. You haven't used the fact that you are at a local maximum and that f"= 0 there. The Taylor's series for f(x), about x= a, is f(a)+ f'(a)(x-a)+ f"(a)/2 (x-a)²+ (f'''(a)/6) (x-a)³+ higher order terms. If x= a is a local maximum then its first derivative, f'(a), is 0 and you are told that the second derivative, f"(a) is also 0: f(x)= f(a)+ (f'''(a)/6)(x-a)³+ higher order terms. For x sufficiently near a, we can ignore those "higher order terms" in comparison to f(a) and (f'''(a)/6) (x-a)³. Think about what happens if x is slightly larger than a and what happens if x is slightly larger than a.

 

[sdff22]

You should know more than that. You haven't used the fact that you are at a local maximum and that f"= 0 there. The Taylor's series for f(x), about x= a, is f(a)+ f'(a)(x-a)+ f"(a)/2 (x-a)²+ (f'''(a)/6) (x-a)³+ higher order terms. If x= a is a local maximum then its first derivative, f'(a), is 0 and you are told that the second derivative, f"(a) is also 0: f(x)= f(a)+ (f'''(a)/6)(x-a)³+ higher order terms. For x sufficiently near a, we can ignore those "higher order terms" in comparison to f(a) and (f'''(a)/6) (x-a)³. Think about what happens if x is slightly larger than a and what happens if x is slightly larger than a.

Thank you very much for he help.
So should I consider different cases like:
1. x-->a+ (i.e. x-a>0), then if f(x)>f(a)==>f"'(a)>0
if f(x)<f(a)==>f"'(a)<0

and the same argument for x-->a- (i.e. x-a<0)

but I am confused about this fact that having both first and second derivative equal to zero tells me that when x gets very close to a, then f(x) gets very close or equal to f(a). Knowing this, doesn't mean that f"'(a)=0 using the Taylor expansion of f(x) you have written?

Thanks again,

 

[HallsofIvy]

No, it doesn't. It means that f(x)= f(a)+ f'''(a)/6 (x- a)³+ higher power terms. Putting x= a tells you nothing about f'''(a) because the (x-a)³ term is 0.

Now, if x is slightly larger than a, what is the sign of (x-a)³? What if x is slightly smaller than a?