Find the minumum and Maximum value of

May 2010
3
1
Varanasi, Uttar Pradesh-India
2^sin x + 2^cos x......where ^ is symbol for 'to the power'
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
Can you solve where \(\displaystyle (2^{sin x} + 2^{cos x})'=0\) ??
 
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May 2010
3
1
Varanasi, Uttar Pradesh-India
a..certainly I can I can use differential and..but like if I take the 2 numbers in A.P.....then by using A.P.>GP...
\(\displaystyle 2^{sin x}+2^{cos x} >=2^{1-1/sqrt(2)}\)

so I am able to get the minimum value...but is there an algebraic way to solve it...not using calculus...I am rather uncomfortable with calculus if it goes complex.
 
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Jun 2009
806
275
a..certainly I can I can use differential and..but like if I take the 2 numbers in A.P.....then by using A.P.>GP...
\(\displaystyle 2^{sin x}+2^{cos x} >=2^{1-1/sqrt(2)}\)

so I am able to get the minimum value...but is there an algebraic way to solve it...not using calculus...I am rather uncomfortable with calculus if it goes complex.
sin(x) and cos(x) are equal when x = π/4. The value is \(\displaystyle \frac{1}{\sqrt{2}\)

sin(x) and cos(x) are equal but having opposite sign when x = 3π/4.

Hence maximum value of \(\displaystyle 2^{sin(x)} + 2^{cos(x)} = 2\times2^\frac{1}{\sqrt{2}}\)

Minimum value is \(\displaystyle 2^\frac{1}{\sqrt{2}} + 2^\frac{-1}{\sqrt{2}}\)
 
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