One-to-one functions (C3) Watch

Eau
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#1
Report Thread starter 12 years ago
#1
How do I show that the function

f(x)=e^(sinx), where x E [-pi/2, pi/2]

is a one-to-one function?
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AlphaNumeric
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#2
Report 12 years ago
#2
sin(x) in that range is monotonically increasing, so e^(sinx) is monotonically increasing, and therefore is 1-1.
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latentcorpse
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#3
Report 12 years ago
#3
plot it and then trace a horizontal line over it and it should only ever touch at one point - within the domain given sine is always one-to-one and so is e^x therefore when u sub sin x into e^x to give e^sinx u still have a one-to-one function
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