Question #143f6

1 Answer
Dec 14, 2017

The range is the closed interval #=[2sinh(-1),2sinh(2)]=[(1-e^{2})/e,(e^{4}-1)/(e^{2})]approx [-2.3504,7.2537]#.

Explanation:

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

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

Clearly #f'(x)>0# for all #x\in RR#, implying that #f(x)# is strictly increasing over #RR#, and also over #[-1,2]#.

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

This implies that the range is the closed interval #=[2sinh(-1),2sinh(2)]=[(1-e^{2})/e,(e^{4}-1)/(e^{2})]approx [-2.3504,7.2537]#.