My Math Forum Range of function

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 April 8th, 2013, 07:30 PM #1 Senior Member   Joined: Feb 2013 Posts: 114 Thanks: 0 Range of function Find the range of function f(x) = $sin^2x-sinx+1$ This function can be written as $(sinx-\frac{1}{2})^2+\frac{3}{4}$............(i) We know the range of sinx function is [-1,1] therefore we can write : $-1 \leq sinx \leq 1 \Rightarrow -1 -\frac{1}{2} \leq sinx -\frac{1}{2} \leq 1 -\frac{1}{2}$ $\frac{-3}{2} \leq sinx -\frac[1}{2} \leq \frac{1}{2}$ We can write this further as : $(-\frac{3}{2})^2 \leq (sinx -\frac{1}{2})^2 \leq (\frac{1}{2})^2$ which is equal to : $\frac{9}{4} \leq (sinx -\frac{1}{2})^2 \leq \frac{1}{4}$ Also $\frac{9}{4} +\frac{3}{4} \leq (sinx -\frac{1}{2})^2 + \frac{3}{4}\leq \frac{1}{4} +\frac{3}{4}$ Will it give the right range.... I am not getting it correct ...please suggest
April 8th, 2013, 08:20 PM   #2
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Joined: Dec 2006
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Re: Range of function

Hello, sachinrajsharma!

Quote:
 $\text{Find the range of function: }\;f(x) \:=\:\sin^2x\,-\,sin x\,+\,1$ $\text{This function can be written as: }\:f(x) \:=\: \left(\sin x\,-\,\frac{1}{2}\right)^2\,+\,\frac{3}{4}$

$\text{Note that a square of a quantity is never negative.}
\text{Hence: }\:\left(\sin x\,-\,\frac{1}{2}\right)^2\:\ge\:0 \;\;\;\Rightarrow\;\;\;\left(\sin x\,-\,\frac{1}{2}\right)^2\,+\,\frac{3}{4} \;\ge\;\frac{3}{4}$

$\text{The maximum value occurs when }x= \frac{3\pi}{2}$

$\;\;\;\left(\sin\frac{3\pi}{2}\,-\,\frac{1}{2}\right)^2\,+\,\frac{3}{4} \;=\;\left(-1\,-\,\frac{1}{2}\right)^2\,+\,\frac{3}{4} \;=\;\left(-\frac{3}{2}\right)^2\,+\,\frac{3}{4} \;=\;\frac{9}{4}\,+\,\frac{3}{4} \;=\;\frac{12}{4} \;=\;3$

$\text{The range is: }\:\frac{3}{4}\;\le\:f(x)\;<\;3$

 April 8th, 2013, 10:09 PM #3 Senior Member   Joined: Feb 2013 Posts: 114 Thanks: 0 Re: Range of function thanks ...but it should be $\frac{3}{4} \leq f(x) \leq 3$

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