Results 1 to 4 of 4

Math Help - [SOLVED] Domain (and range) of a logarithmic function

  1. #1
    Member courteous's Avatar
    Joined
    Aug 2008
    From
    big slice of heaven
    Posts
    206

    Question [SOLVED] Domain (and range) of a logarithmic function

    f(x)=\cos{x}-\sin{x}. What are domain and range of g(x)=log_2f(x)?

    Altering f(x) so that it contains only \sin (first part of the problem): f(x)=\cos{x}-\sin{x}=\sin{(\frac{\pi}{2}-x)}-\sin{x}=...=-\sqrt{2}\sin{(x-\frac{\pi}{4})} (this is assuredly correct).

    Now, for the logarithm ( g(x)) and its range, I have no hunch of proper solution (for domain, sense tells me that the solution is where f(x)>0 ( x\in(-\frac{5\pi}{4}+2k\pi, \frac{\pi}{4}+2k\pi)), but I assume that is NOT rigour enough, is it?) ...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    Quote Originally Posted by courteous View Post
    f(x)=\cos{x}-\sin{x}. What are domain and range of g(x)=log_2f(x)?

    Altering f(x) so that it contains only \sin (first part of the problem): f(x)=\cos{x}-\sin{x}=\sin{(\frac{\pi}{2}-x)}-\sin{x}=...=-\sqrt{2}\sin{(x-\frac{\pi}{4})} (this is assuredly correct).

    Now, for the logarithm ( g(x)) and its range, I have no hunch of proper solution (for domain, sense tells me that the solution is where f(x)>0 ( x\in(-\frac{5\pi}{4}+2k\pi, \frac{\pi}{4}+2k\pi)), but I assume that is NOT rigour enough, is it?) ...

    your domain is correct. and the rage is easier. the maximum value of f(x) is \sqrt{2} as the log function is an increasing function the maximum value of g(x) is log_{2} (\sqrt{2}) = \frac{1}{2}. so the range will be form minus infinity to a half.

    Bobak
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member courteous's Avatar
    Joined
    Aug 2008
    From
    big slice of heaven
    Posts
    206

    Rigour.

    How would you mathematically (rigorously) write down "where f(x) is max, g(x) is max. Therefore, range of g(x) is..." (same for domain; I only got the solution, but how should one formalize the steps to get the solution?)

    Extra question: f(x)>0 \Longleftrightarrow x\in(-\frac{5\pi}{4}+2k\pi, \frac{\pi}{4}+2k\pi) Is this 'right-to-left arrow' correctly put (is the sentence mathematically correct)? Or should there be only 'right arrow' ( \Longrightarrow)? Or any other arrow?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    Quote Originally Posted by courteous View Post
    How would you mathematically (rigorously) write down "where f(x) is max, g(x) is max. Therefore, range of g(x) is..." (same for domain; I only got the solution, but how should one formalize the steps to get the solution?)

    Extra question: f(x)>0 \Longleftrightarrow x\in(-\frac{5\pi}{4}+2k\pi, \frac{\pi}{4}+2k\pi) Is this 'right-to-left arrow' correctly put (is the sentence mathematically correct)? Or should there be only 'right arrow' ( \Longrightarrow)? Or any other arrow?
    g(x)=log_2f(x)

    g'(x)= \frac{f'(x)}{\ln 2 f(x)}

    g''(x)= \frac{f''(x)f(x) - [f'(x)]^2}{(\ln 2 f(x))^2}

    it should be clear that g'(x) is only zero if f'(x), so this shows the stationary points on g occur at the same points as f. However you must now show that they are of the same nature. when f is at a maximum f''(x) is negative and f'(x) is zero. so at a maximum point of f we have...

    g''(x)= \frac{f''(x)f(x)}{(\ln 2 f(x))^2}

    so g''(x) is negative if f''(x) is negative and f(x) is positive. and f(x) >0 is required for g to be defined. I guess this is a more rigourous method of showing they have maximums at the same point.


    Also your use of the right to left arrow is correct, as the statement read correctly both ways. as if x is a member of the open interval (-\frac{5\pi}{4}+2k\pi, \frac{\pi}{4}+2k\pi) then f is positive, and also if f(x) is positive then x must be a member of the open interval.

    Bobak
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Domain and Range of this Function
    Posted in the Algebra Forum
    Replies: 6
    Last Post: May 7th 2010, 05:33 AM
  2. Replies: 2
    Last Post: September 30th 2009, 04:29 AM
  3. [SOLVED] domain and range
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: March 13th 2009, 02:15 PM
  4. Replies: 3
    Last Post: January 28th 2009, 10:42 PM
  5. [SOLVED] Domain and range
    Posted in the Algebra Forum
    Replies: 3
    Last Post: August 21st 2008, 09:05 PM

Search Tags


/mathhelpforum @mathhelpforum