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]}