Consider the function #y = e^x# defined on #[-1, 2]#. What are the maximum and minimum of the function on this interval?

Question

Consider the function #y = e^x# defined on #[-1, 2]#.  What are the maximum and minimum of the function on this interval?

Elijah Abram · Accepted Answer

This function has no absolute maximum or minimum, but would have a local minimum at #y = 1/e# and a local maximum at #y = e^2# on the interval #[-1, 2]#

We don't need calculus to explain this. The graph of any exponential function will have a domain of all the real numbers. The graph of #e^x# will have a range of #y>0#, however the graph will never touch #y=0#, therefore #y = 0# cannot be considered a minimum.

In calculus speak, we would say #lim_(x-> -oo) e^x = 0#, which means that the function approaches #y = 0# as #x# approaches negative infinity.

So the local max/min will be the two end points of our closed interval. Hence there willl be a local maximum at #x = 2# and local minimum at #x = -1#

Hopefully this helps!

Elijah Abram · Answer

undefined The maximum value of the function $ y = e^x $ on the interval $[-1, 2]$ occurs at $x = 2$, where $y = e^2$. The minimum value of the function on the same interval occurs at $x = -1$, where $y = e^{-1}$. Therefore, the maximum value is $e^2$ and the minimum value is $e^{-1}$.