How do you sketch #y=e^absx#?

1 Answer
Feb 4, 2017

See explanation...

Explanation:

When #x = 0# the given formula gives #y = e^abs(color(blue)(0)) = 1#

Also note that since #y# is a function of #abs(x)#, its graph will be symmetrical about the #y# axis.

The function #f(x) = e^x# is well known for being equal to its own derivative. That is, the instantaneous slope at any point is itself #e^x#.

In symbols we can write:

#d/(dx) e^x = e^x#

So we find:

#lim_(x->0) d/(dx) e^x = lim_(x->0) e^x = e^0 = 1#

Hence the slope of our given function approaches #1# as #x->0^+#, that is as #x->0# from the right.

As a result, the graph of #y = e^abs(x)# forms a V shape at #(0, 1)# with slope #+-1#

We can also evaluate #y=e^abs(x)# for a few values of #x# to provide some points through which the graph passes:

#e^abs(1) ~~ 2.718#

#e^abs(2) ~~ 7.389#

Putting this information together we can plot the graph of #y = e^abs(x)# as a kind of steep U shape with a pointed vertex:
graph{(y-e^abs(x))(x^2+(y-1)^2-0.02)((x-1)^2+(y-e)^2-0.02)((x-2)^2+(y-e^2)^2-0.02) = 0 [-10.375, 9.625, -1.32, 8.68]}