How do you sketch y=e^absxy=e|x|?

1 Answer
Feb 4, 2017

See explanation...

Explanation:

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

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

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

In symbols we can write:

d/(dx) e^x = e^xddxex=ex

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]}