To prove this equation can exist or not hmm

[BrendanM]

The question is : Find an F(x) where fprime(-1) = 1/2 , fprime(0) = 0 and fdoubleprime>0 if it doesn't exist prove why.

I cannnot explain it that well but here i go, i feel there is no equation that can be made for this. For an equation for be concave up on all intervals it must be in the form Ax^multipleof2 + B^multipleof2 + C^multipleof2 + etc etc etc... + Dx + E, where A and B and C... are positive constants I say this because f double prime will onyl be positive for all x if the second dirivitive has only positive constants or terms of x^multipleof2. Then the first dirivitive will leave you with x^odd number + a constant, f(-1) must be equal to 1/2, so it must be positive since x^oddnumber will be negative. but then when it says fprime(0) = 0 this cannot be because you will have 0^oddnubmer + constant.

blah that's as far as i could take it I am not sure if it makes sense.. but can someone help me on the correct path.

 

[shmoe]

-not all functions are polynomials
-a polynomial can be concave up everywhere and still have negative coefficients or odd powers, x^{2}-x-1 does all this
-latex is a beautiful thing and will make your posts legible with little effort, see https://www.physicsforums.com/misc/howtolatex.pdf

For your question, if f''(x)>0 for all x what can you say about f'(x)? Is it increasing? Decreasing? Neither?

 

[HallsofIvy]

Since f'' is just the derivative of f', this exactly the same as:

"Does there exist a function f, whose derivative is always positive, such that f(-1)= 1/2 and f(0)= 0". Since f ' is always positive, what does that tell you about f?

 

[philosophking]

I think HallsofIvy meant to put primes in front of those f's, but anyway... :)

Think of it like this: The slope of a function is positive at -1, then it turns to zero at 0. Draw that out. What would a function like that look like? Increasing then not. So does this mean f can be concave up for all x?