# Thread: [SOLVED] Domain (and range) of a logarithmic function

1. ## [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?) ...

2. Originally Posted by courteous
» $f(x)=\cos{x}-\sin{x}$. 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

3. ## 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?

4. Originally Posted by courteous
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