Question:
Find the range of following function:
\(y=\sin^2{x} - 4\sin{x} +7\)
Solution:
We are given, \(y=\sin^2{x}-4\sin{x}+7\)
First of all, we will convert it in a perfect square.
So,
\(y=(\sin{x})^2-2(2)(\sin{x})+2^2+3\)
\(\implies y=(\sin{x}-2)^2+3\)
We know, the range of \(\sin{x}\) is \([-1,1]\).
So, we will have:
At \(-1\),
\(y=(-1-2)^2+3\)
\(\implies y=9+3=12\)
At \(1\),
\(y=(1-2)^2+3\)
\(\implies y=1+3=4\)
Also, we know that \(\textbf{iff} \sin{x}-2=0\) then we can get the lesser value than \(4\) which is at \(1\), but this is not possible for any value of \(x\) because we know the range of \(\sin{x}\).
Therefore, \(y_{\text{max}}=12\)
And, \(y_{\text{min}}=4\)
Range of \(y\) is \([4,12]\).
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