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Does this need Taylor?



I think it probably does but I have no clue how to go about it!
Reply 1
Avatar for nuodai
nuodai
17
I don't think so. You can rearrange the inequality to be 15x−x15>−45\dfrac{1}{5}x - x^{\frac{1}{5}} > -\dfrac{4}{5} for all 0<x<10 < x < 1. This indicates that you might want to find the minimum value that 15x−x15\dfrac{1}{5}x-x^{\frac{1}{5}} takes in the interval 0<x<10<x<1 -- look at the extreme values, and differentiate to see if there are any stationary points in the interval (if not, you can safely assume that the graph is either increasing or decreasing and so achieves its max/min values at x=0 or x=1).
Reply 2
Avatar for Dagnabbit
Dagnabbit
OP
14
Original post by nuodai
I don't think so. You can rearrange the inequality to be 15x−x15>−45\dfrac{1}{5}x - x^{\frac{1}{5}} > -\dfrac{4}{5} for all 0<x<10 < x < 1. This indicates that you might want to find the minimum value that 15x−x15\dfrac{1}{5}x-x^{\frac{1}{5}} takes in the interval 0<x<10<x<1 -- look at the extreme values, and differentiate to see if there are any stationary points in the interval (if not, you can safely assume that the graph is either increasing or decreasing and so achieves its max/min values at x=0 or x=1).


Does this have anything to do with the Mean Value Theorem? The previous parts of this question were MVT.
Reply 3
Avatar for nuodai
nuodai
17
Original post by Dagnabbit
Does this have anything to do with the Mean Value Theorem? The previous parts of this question were MVT.


Not that I can see.

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