# Question #143f6

Dec 14, 2017

The range is the closed interval $= \left[2 \sinh \left(- 1\right) , 2 \sinh \left(2\right)\right] = \left[\frac{1 - {e}^{2}}{e} , \frac{{e}^{4} - 1}{{e}^{2}}\right] \approx \left[- 2.3504 , 7.2537\right]$.

#### Explanation:

Let $f \left(x\right) = {e}^{x} - {e}^{- x} = 2 \sinh \left(x\right)$, where $\sinh \left(x\right)$ is the hyperbolic sine function .

Then $f ' \left(x\right) = {e}^{x} + {e}^{- x} = 2 \cosh \left(x\right)$, where $\cosh \left(x\right)$ is the hyperbolic cosine function.

Clearly $f ' \left(x\right) > 0$ for all $x \setminus \in \mathbb{R}$, implying that $f \left(x\right)$ is strictly increasing over $\mathbb{R}$, and also over $\left[- 1 , 2\right]$.

Therefore, the minimum value of $f$ over $\left[- 1 , 2\right]$ is $f \left(- 1\right) = 2 \sinh \left(- 1\right) = {e}^{- 1} - {e}^{1} = \frac{1 - {e}^{2}}{e} \approx - 2.3504$ and the maximum value of $f$ over $\left[- 1 , 2\right]$ is $f \left(2\right) = 2 \sinh \left(2\right) = {e}^{2} - {e}^{- 2} = \frac{{e}^{4} - 1}{{e}^{2}} \approx 7.2537$.

This implies that the range is the closed interval $= \left[2 \sinh \left(- 1\right) , 2 \sinh \left(2\right)\right] = \left[\frac{1 - {e}^{2}}{e} , \frac{{e}^{4} - 1}{{e}^{2}}\right] \approx \left[- 2.3504 , 7.2537\right]$.