Question #10930
the minimum value of 2^(sinx-1) + 2^(cosx-1) is
Expert's answer
This function is always positive (it's obvious). Now we can put 1/2 aside and concentrate on expression 2^sinx+2^cosx. Let's find it's derivative:
f'(x)=(cosx*2^sinx)/ln2-(sinx*2^cosx)/ln2=0. 2^sinx*cosx=2^cosx*sinx it's possible when cosx=sinx. As we're searching the minimum we interested in such
values of x when cosx and sinx are both negative. So x=5pi/4+kpi, where k is
random integer. The value of our expression in this case is 2^(1/sqrt(2)).
f'(x)=(cosx*2^sinx)/ln2-(sinx*2^cosx)/ln2=0. 2^sinx*cosx=2^cosx*sinx it's possible when cosx=sinx. As we're searching the minimum we interested in such
values of x when cosx and sinx are both negative. So x=5pi/4+kpi, where k is
random integer. The value of our expression in this case is 2^(1/sqrt(2)).
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