Mean Value Theorem: demonstrate inequality b^n - a^n > nb^(n-1) (b - a), if b > a
- Thread starter Brunorp
- Start date
- Joined
- Feb 4, 2004
- Messages
- 16,550
In order to apply the Mean Value Theorem (here), you need to start with a defined interval. Did the exercise provide any information in addition to what is posted above?
Also, the Theorem usually leads to a fractional form which, in this case, I suspect will be something along the lines of the following:
. . . . .\(\displaystyle \dfrac{b^n\, -\, a^n}{b\, -\, a}\)
Does this suggest anything to you (like this, maybe)?
In order to apply the Mean Value Theorem (here), you need to start with a defined interval. Did the exercise provide any information in addition to what is posted above?
Also, the Theorem usually leads to a fractional form which, in this case, I suspect will be something along the lines of the following:
. . . . .\(\displaystyle \dfrac{b^n\, -\, a^n}{b\, -\, a}\)
Does this suggest anything to you (like this, maybe)?
I noticed that the inequality given leads to the fractional form as the Theorem, but I'm not sure about the derivative's sign. Considering the function f(x) = xˆn , if a < c < b, could I afirm that f'(a) < f'(c) < f'(b)? Because that's exactly the inequality to which the exercise leads, isn't it?
- Joined
- Feb 4, 2004
- Messages
- 16,550
With the current givens, no. Consider f(x) = x^3, and x < 0. Also, consider the "solution" posted for (6) here, which appears to prove the exact opposite of the inequality that you've posted.
Does the exercise provide any other information? Thank you!
With the current givens, no. Consider f(x) = x^3, and x < 0. Also, consider the "solution" posted for (6) here, which appears to prove the exact opposite of the inequality that you've posted.
Does the exercise provide any other information? Thank you!
No, there's no further information. I believe you're right. Thanks for the help!