# Math Help - Functions and derivatives. Any experts out there?

1. ## Functions and derivatives. Any experts out there?

I'm confused with this whole concept of functions and derivatives. If you could help me that'd be wonderful. Thank you so much!

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Suppose that f and f ' are defined for all values of x...

f' has the following properties:
f' (x) > 0 for all x
f' has a maximum value of 5. This value occurs at the approximate x-coordinate 0.347
As x increases towards infinity, f ' (x) decreases monotonically toward 0.
As x decreases towards negative infinity, f ' (x) decreases monotonically toward 0.
Also, f (x) =2.5

1) explain why f (x) could not be the function ax^5 + bx^4 + cx^3 + dx^2 + ex + f for any values of A, B, C, D, E, and F.
2) Use the "speed limit principle" to show that the value of f(1) must be less than 6.
3) Use the "speed limit principle" to find upper and lower bounds on the value of f(0).

2. Originally Posted by lilxswttooth
I'm confused with this whole concept of functions and derivatives. If you could help me that'd be wonderful. Thank you so much!

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Suppose that f and f ' are defined for all values of x...

f' has the following properties:
f' (x) > 0 for all x (*)
f' has a maximum value of 5. This value occurs at the approximate x-coordinate 0.347
As x increases towards infinity, f ' (x) decreases monotonically toward 0.
As x decreases towards negative infinity, f ' (x) decreases monotonically toward 0. (**)
Also, f (x) =2.5

1) explain why f (x) could not be the function ax^5 + bx^4 + cx^3 + dx^2 + ex + f for any values of A, B, C, D, E, and F.
2) Use the "speed limit principle" to show that the value of f(1) must be less than 6.
3) Use the "speed limit principle" to find upper and lower bounds on the value of f(0).
1) Let $f(x):=ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f$ for $x\in\mathbb{R}$. Clearly, a polynomial (unless identically $0$) tends to $\pm\infty$ at $\pm\infty$. Since $f^{\prime}$ (which is also a polynomial) tends to $0$ at $-\infty$ by (**), we must have $f^{\prime}\equiv0$, and contradicts to (*).