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.#