Question

f(x) = ln(4 − ln x)
find asymptotes, increasing and decreasing intervals min and max number, concave up concave down

Answer 1

the example problem was ln(1-lnx) which makes it easy to find the asymtote because ln e =1 but how can do the same thing with the numebr 4?

Answer 2

\(\ln x\) is defined for \(x>0\), so right away you know that \(y=0\) is a vertical asymptote.

For the whole function \(f(x)\), you have it defined for \(4-\ln x>0\), or \(\ln x<4\), or \(x<e^4\). This means \(y=e^4\) is another vertical asymptote.

Outside of the interval \(\left(0,e^4\right)\), \(f(x)\) is not defined.

Answer 3

awesome

Answer 4

I figure'd I'd give you a hand with the derivatives as well...\[\begin{align*}f'(x)&=\frac{\dfrac{d}{dx}[4-\ln x]}{4-\ln x}\\\\&=\frac{-\dfrac{1}{x}}{4-\ln x}\\\\&=\frac{1}{x(\ln x-4)}\end{align*}\]
\[\begin{align*}f''(x)&=\frac{-\dfrac{d}{dx}[x(\ln x-4)]}{x^2(\ln x-4)^2}\\\\
&=\frac{\dfrac{d}{dx}[x(4-\ln x)]}{x^2(\ln x-4)^2}\\\\
&=\frac{(4-\ln x)+x\dfrac{\dfrac{d}{dx}[4-\ln x]}{4-\ln x}}{x^2(\ln x-4)^2}\\\\
&=\frac{4-\ln x-\dfrac{1}{4-\ln x}}{x^2(4-\ln x)^2}\\\\
&=\frac{(4-\ln x)^2-1}{x^2(4-\ln x)}
\end{align*}\]

Answer 5

That second derivative might not be right. Hold on...

Answer 6

It wasn't!\[\begin{align*}f''(x)&=\frac{-\dfrac{d}{dx}[x(\ln x-4)]}{x^2(\ln x-4)^2}\\\\
&=\frac{\dfrac{d}{dx}[x(4-\ln x)]}{x^2(\ln x-4)^2}\\\\
&=\frac{(4-\ln x)+x\dfrac{d}{dx}[4-\ln x]}{x^2(\ln x-4)^2}\\\\
&=\frac{(4-\ln x)+x\left(-\dfrac{1}{x}\right)}{x^2(\ln x-4)^2}\\\\
&=\frac{3-\ln x}{x^2(\ln x-4)^2}\\\\
\end{align*}\]

Answer 7

cool thats what i got

Answer 8

so out of the second derivative i get e^3, e^4? so my intervals are 0,e^3 e^3,e^4

Answer 9

The possible inflection points are \(x=0,e^3,e^4\), though we know the first and third don't apply because they're the endpoints of the domain. Your intervals are correct, assuming you mean \((0,e^3)\) and \((e^3,e^4)\).