Question #ef937

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Answer 1 · 2017-09-04T10:59:57

$e^(x-1) + 1$

Could also write this as $(e^x + e)/e$ if that is more helpful.

$(e^x(1+e^(1-x)))/e^1$

Divide the terms outside the bracket (ie subtract indices)

$=e^(x-1)(1 + e^(1-x))$

Now multiply through the bracket (ie add indices)

$= e^(x-1)*1 + e^(x-1)*e^(1-x)$

$=e^(x-1) + e^0 = e^(x-1) + 1$

Answer 2 · 2017-09-04T11:11:15

$e^(x-1)+1.$

Assuming that, simplification for the Exression ${e^x(1+e^(1-x))}/e$

is reqd.

$"The Exp.="{e^x(1+e^(1-x))}/e$

$=(e^x/e^1)(1+e^(1-x)),$

$=e^(x-1)*(1+e^(1-x)).............[because, a^m/a^n=a^(m-n)],$

$=e^(x-1)*1+e^(x-1)*e^(1-x)........[because, l(m+n)=lm+ln],$

$=e^(x-1)+e^{(x-)+(1-x)}......[because, a^m*a^n=a^(m+n)],$

$=e^(x-1)+e^0,$

$:." The Exp.="e^(x-1)+1, because, a^0=1, ane0.$