There exists a function f such that f(x) < 0, f'(x) < 0 and f''(x) > 0 for all real x. True or false?

Question

Please help with this too:

There exists a function f such that f(x) > 0, f'(x)<0 and f''(x)>0 for all real x. True or false?

Thanks a lot! :)

Answer 1

The first one's false. Proof: Suppose f is such a function. Then f'(0) < 0. Suppose ∃x<0 s.t. f'(x)>f'(0). Then by the mean value theorem, ∃c∈(x, 0) s.t. f''(c) = (f'(0)-f'(x))/(0-x). Since f'(x)>f'(0), the numerator is negative, and we have stipulated that x<0, so the denominator is positive. Therefore, (f'(0)-f'(x))/(0-x) < 0, contradicting the fact that f''(x) is always positive. So ∀x≤0, f'(x)≤f'(0).

Now, if there were some x<0 such that f(x) -xf'(0), and thus (f(0)-f(x))/(-x) > f'(0) (we do not reverse the inequality, because -x is positive). But per the mean value theorem, ∃c∈(x, 0) s.t. f'(c) = (f(0)-f(x))/(0-x), and so we would have c<0 s.t. f'(c) > f'(0), contradicting the previous result. It follows that ∀x≤0, f(x) ≥ f(0) + xf'(0)

Now, having established this, we note that since f'(0) < 0, [x→-∞]lim f(0) + xf'(0) = ∞, so in particular there exists x<0 such that f(0) + xf'(0)>0, and so there would have to exist x<0 such that f(x) > 0, contradicting the assumption that f(x)<0. Therefore, no function with f(x)<0, f'(x)<0, and f''(x)>0 can be defined on the whole real line. Q.E.D.

The second one is much easier to solve: Yes, such a function does exist -- one example is f(x)=e^(-x).

Answer 2

False.
F(x)<0 tells you that the regular function is below the x-axis. Too bad this isn't too much help, as this could mean anything :P
F'(x)<0 tells you that F(x) has a negative slope, which means that F(x) is going downwards further
But, F"(x)>0 tells you that the concavity of the function is positive, which means that the function has to be going up at at least a gemoetric progression. You cannot have the function to keep getting lower while having a positive concavity, therefore the statement is false.

Answer 3

Both are true. the lower limit of the range of the function will be the minima in both the cases and thus both the statements are true.
In the first one the function is above x axis but with negative slope so the value keeps decreasing and it is upto a point so that point is the minima and so f''(x) gives a positive value which means a minima exists.
In the second one it is similar except that the function is below x axis.. here also minima exists at the lowermost point which is the lower limit of the range of the function.

Answer 4

This has to do with the Intermediate fee Theorem. it truly is only actual if the function is non-end. because of the fact the assertion above does no longer assure continuity, the assertion is fake. Edit: it relatively is not the advise fee Theorem (MVT).

Answer 5

true I hate logics