Find critical numbers and absolute extrema I hope the picture is clear and I need help?

enter image source here

1 Answer
Mar 3, 2018

Please see below.

Explanation:

We have #f(x)=e^x/(x+1)#. to identfy critical numbers, we must first take the derivative of the function and then set it equal to #0# and solve for #x#.

Here #f'(x)=(e^x(x+1)-e^x)/(x+1)^2=(xe^x)/(x+1)^2#

and #f'(x)=0#, when #x=0# and when #x->-oo#

at these values of #f(x)# is #1# and #0#. Hence critical values are at #(0,1)# and #-oo,0)#.

Further as #x->-1#, #f(x)->+-oo#, hence we have a vertical asymptote at #x=-1#

We also have #f''(x)=((x+1)^2(e^x+xe^x)-2xe^x(x+1))/(x+1)^4#

= #((x+1)(e^x+xe^x)-2xe^x)/(x+1)^3#

andat #x=0#, we have a minima as #f''(0)=-1/8<0# in the interval #[-1/2,2]#,

graph{e^x/(x+1) [-10, 10, -5, 5]}