What are the absolute extrema of $f (x) = | sin (x) + ln (x) |$ $f(x)= |sin(x) + ln(x)|$ on the interval (0 ,9]?

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Answer 1 · 2017-05-21T17:35:44

No maximum. Minimum is $0$ $0$ .

No maximum
As $x \to 0$ $xrarr0$ , $sin x \to 0$ $sinxrarr0$ and $ln x \to - \infty$ $lnxrarr-oo$ , so

$lim_(xrarr0) abs(sinx+lnx) = oo$

So there is no maximum.

No minimum

Let $g(x) = sinx+lnx$ and note that $g$ is continuous on $[a,b]$ for any positive $a$ and $b$ .

$g(1) = sin1 > 0$ $" "$ and $" "$ $g(e^-2) = sin(e^-2) -2 < 0$ .

$g$ is continuous on $[e^-2,1]$ which is a subset of $(0,9]$ .

By the intermediate value theorem, $g$ has a zero in $[e^-2,1]$ which is a subset of $(0,9]$ .

The same number is a zero for $f(x) = abs(sinx+lnx)$ (which must be non-negative fo all $x$ in the domain.)

What are the absolute extrema of f(x)= |sin(x) + ln(x)|f(x)=|sin(x)+ln(x)| f(x)= |sin(x) + ln(x)| on the interval (0 ,9]?