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

##### 1 Answer

See explanation...

#### Explanation:

When

Also note that since

The function

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

As a result, the graph of

We can also evaluate

#e^abs(1) ~~ 2.718#

#e^abs(2) ~~ 7.389#

Putting this information together we can plot the graph of

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]}