Question #45820

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Question #45820

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Answer 1 · 2017-11-22T11:13:36

$(f \cdot g) (x) = \frac{log (x + 1)}{x - 2}$ $(f*g)(x)=log(x+1)/(x-2)$ , $x$ $x$ $\in (- 1, 2) \cup (2, + \infty)$ $in(-1,2)uu(2,+oo)$

Explanation:

$\cdot$ $*$ $f (x) = \frac{1}{x - 2}$ $f(x)=1/(x-2)$

$x - 2 \neq 0$ $x-2!=0$ $\Leftrightarrow$ $<=>$ $x \neq 2$ $x!=2$

so $D f = R - {2}$ $Df=R-{2}$ = $(- \infty, 2) \cup (2, + \infty)$ $(-oo,2)uu(2,+oo)$

$\cdot$ $*$ $g (x) = log (x + 1)$ $g(x)=log(x+1)$

$x + 1 > 0$ $x+1>0$ $\Leftrightarrow x > - 1$ $<=>x> -1$
so $D g = (- 1, + \infty)$ $Dg=(-1,+oo)$

For $x$ $x$ $\in$ $in$ $D f \cap D g$ $DfnnDg$ $=$ $=$ $(- 1, 2) \cup (2, + \infty)$ $(-1,2)uu(2,+oo)$
$(f \cdot g) (x)$ $(f*g)(x)$ $=$ $=$ $f (x) \cdot g (x) = \frac{1}{x - 2} \cdot log (x + 1)$ $f(x)*g(x)=1/(x-2)*log(x+1)$ $= \frac{log (x + 1)}{x - 2}$ $=log(x+1)/(x-2)$

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Question #45820

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