function value

[Valentas]

Well I stucked with these:

a) \(\displaystyle f(x) = 2^{1-3sin^{2}x}\)

c) \(\displaystyle f(x) = 2^{|cosx|}\)

I need to get E =

 

[stapel]

Well I stucked with these:

a) \(\displaystyle f(x) = 2^{1-3sin^{2}x}\)

c) \(\displaystyle f(x) = 2^{|cosx|}\)

I need to get E =

There is no "E" in your function statements (and these really aren't "equations" that one would "solve"), so I'm afraid I don't understand what you're getting at...? :wink:

 

[Valentas]

Well, I might don't know how to tell that thing in english but I need to find an interval. e.g. sinx value domain is {-1;1} . I don't know how to solve that a and c ..

e.g. the answer of a) is {0.25;2}

 

[soroban]

Hello, Valentas!


The problem asks for the range of the function.

\(\displaystyle a)\;f(x) \,=\, 2^{1-3\sin^2x}\)

\(\displaystyle \begin{array}{ccccccc}\text{We have:} & \text{-}1 \:\le\:\sin x \:\le\:1 \\ \\ \text{Square:} & 0 \:\le\:\sin^2\!x\:\le\:1 \\ \\ \text{Multiply by -3:} & 0 \:\ge\:\text{-}\sin^2\!x\:\ge\:\text{-}3 \\ \\ \text{That is:} & \text{-}3 \:\le\:\text{-}\sin^2\!x\:\le\:0 \\ \\ \text{Add 1:} & \text{-}2 \:\le\:1-\sin^2\!x\:\le\: 1 \\ \\ \text{Exponentiate:} & 2^{\text{-}2} \:\le\:2^{1-3\sin^2\!x} \:\le\:2^1 \\ \\ \text{Therefore: } & \frac{1}{4} \:\le\:f(x) \:\le \: 2 \end{array}\)


\(\displaystyle \text{The range of }f(x)\text{ is: }\:[0.25,\:2]\)