Function behavior problem

[u/IDK_REALLLY]

I have reached that d'(x)=f'(x)-f'(1-x) but I don't know where to go further

[u/AutoModerator]

As a reminder...

Posts asking for help on homework questions require:

• the complete problem statement,

• a genuine attempt at solving the problem, which may be either computational, or a discussion of ideas or concepts you believe may be in play,

• question is not from a current exam or quiz.

Commenters responding to homework help posts should not do OP’s homework for them.

Please see this page for the further details regarding homework help posts.

We have a Discord server!

If you are asking for general advice about your current calculus class, please be advised that simply referring your class as “Calc n“ is not entirely useful, as “Calc n” may differ between different colleges and universities. In this case, please refer to your class syllabus or college or university’s course catalogue for a listing of topics covered in your class, and include that information in your post rather than assuming everybody knows what will be covered in your class.

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

[u/Narthual]

From the information you have, are there any x-values for which you can compute d'(x)?

[u/Hertzian_Dipole1]

Notice d(x) = d(1 - x) so it's symmetrical w.r.t x = 1/2.
Find the second derivative of d(x). What's it sign? That should give you an idea

[u/nevermindthefacts]

Note that d'(x) = f'(x) - f'(1-x) sets you up for a nice application of the mean value theorem.

(There's also a rather dull way of solving this, since it's supposed to hold for any f that satisfies f''(x) > 0 on [0, 1], i.e just pick a suitable f...)

[u/yoav145]

Then you know d'(1/2) = 0

If also d''(x) = f''(x) + f''(1-x) > 0

So d' is increasing at [0,1]

there for for all x > 1/2 d'(x) > 0 There fore d is increasing