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

Jun 6, 2017

This function has no absolute maximum or minimum, but would have a local minimum at $y = \frac{1}{e}$ and a local maximum at $y = {e}^{2}$ on the interval $\left[- 1 , 2\right]$

#### Explanation:

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 ${\lim}_{x \to - \infty} {e}^{x} = 0$, 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!