Question #e0971

1 Answer
NickTheTurtle
Jan 3, 2018

#" "e^x\approxx+1# for #x\approx0#
#\thereforee^-0.015\approx-0.015+1=0.985#

Explanation:

We know that #d/dx(e^x)=e^x# and #e^0=1#.

Then, we can find that the tangent line to #e^x# at #x=0# has a slope of #e^0=1# and has a y-intercept at #(0,e^0)=(0,1)#. Thus, the equation of the tangent line is #y=x+1#.

Graph the tangent line and zoom in:
graph{(y-x-1)(y-e^x)=0 [-0.5,0.5,-0.5,1.5]}

It seems that the line #y=e^x# and #y=x+1# are quite close in near #x=0#. In other words, #e^x\approxx+1# when #x\approx0#.

So, #e^-0.015\approx-0.015+1=0.985#, which is quite close to the actual value #e^-0.015=0.9851# (to four decimal places).

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