What is the range of the function $h(x) = ln(x+6)$ ?

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Answer 1 · 2017-11-16T22:36:15

Answer: Using Monotony/continuity & Domain: $h(Dh)=R$

$h(x) = ln(x+6)$ , $x>$ $-6$

$Dh=(-6,+oo)$

$h'(x)=1/(x+6)$ $(x+6)'$ $=1/(x+6)$ $>0$ , $x> -6$
So that means that $h$ is strictly increasing ↑ in $(-6,+oo)$

$h$ is obviously continuous in $(-6,+oo)$ as composition of $h_1$ (x)=x+6 & $h_2$ (x) = $lnx$

$h(Dh)=h($ (-6,+oo) $)$ = ( $lim_(xrarr-6)h(x)$ , $lim_(xrarr+oo)h(x))$ $=(-oo,+oo)$ $=R$

because $⋅$ $lim_(xrarr-6)h(x)$ = $lim_(xrarr-6)ln(x+6)$

$x+6=y$
$xrarr-6$
$yrarr0$

$= lim_(yrarr0)lny$ $=-oo$

$⋅$ $lim_(xrarr+oo)h(x)$ = $lim_(xrarr+oo)ln(x+6)$ $=+oo$

Note: you can also show this using the reverse $h^-1$ function. ( $y=ln(x+6)=>......)$

What is the range of the function h(x) = ln(x+6)h(x) = ln(x+6)?