# Absolute Maximum and Absolute Minimum of $f(x) = x - \ln(2x)$ on $\left[\frac 12, 2\right]$

Find the Absolute Maximum and Absolute Minimum of: $$f(x) = x - \ln(2x),\space \left[\frac 12, 2\right]$$

I got the End Points at: $\left(\frac 12, \frac 12\right)$ and $(2, .61)$

I found a Critical Value at $x = 1$

Finally I got:$$\text{Absolute Max at }(2, .61)$$ $$\text{Absolute Min at }(1, .31)$$

Am I correct?

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Yes, except that you should make clear that $0.61$ and $0.31$ are only approximations, the exact values being $2-\ln 4=2(1-\ln 2)$ and $1-\ln 2$, respectively. –  Brian M. Scott Nov 10 '12 at 0:38
I'm putting in my answers through WebAssign, and when I enter .61 as the Absolute Max and .31 as the Absolute Min it says I got it wrong. Should I just enter the whole decimal up to 5 places? –  dsta Nov 10 '12 at 0:42
I’m not familiar with that particular system; it might want more precision, or if it’s like one that I did use before I retired, it may even want a symbolic answer, like $1-\ln(2)$. –  Brian M. Scott Nov 10 '12 at 0:44