How do you use the first and second derivatives to sketch #y = e^(1-2x)#?

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Answer 1 · 2017-12-15T06:01:09

See argument below.

#y=e^(1-2x)#

#dy/dx = d/dx e^(1-2x)#

#dy/dx = e^(1-2x) * d/dx (1-2x)#

#dy/dx=-2 e^(1-2x) #

Similarly, #(d^2y)/dx^2 = +4e^(1-2x)#

For extrema points of #y, dy/dx =0#

However, in this case, #-2 e^(1-2x) <0 forall x in RR#

Consider, #lim_(x->+oo) -2 e^(1-2x) =0 and lim_(x->+oo) e^(1-2x) =0#

Also consider, #(d^2y)/dx^2 =+4e^(1-2x) >0 forall x in RR#

Hence, it seems reasonable to deduce that #y# tends to its minimum value of #0# as #x# tends to #+oo#

This helps us visualise the graph of #y# below.

The other important point we need for the graph is at #x=0#

Here, #y = e^(1-0) = e#

We see the point #(0,e)# on the graph below.

graph{e^(1-2x) [-5.69, 5.406, -1.722, 3.827]}