Question

Find the range of \[f(x) = {\sin ^2}x - 3\sin x + 2\].

Answer 1

Hint: From the given function using the basic identity of \[{\left( {a - b} \right)^2}\], simplify the expression that is given to us. We get \[f(x)\]as a function which is greater than zero. Thus form the range of the function.

Complete step-by-step answer:

The range of a function is the complete set of all possible resulting values of the dependent variable, after we have substituted the domain. The range is the resulting y-values we get after substituting all the possible x-values. The range of a function is the spread of possible y-values (minimum y-value to maximum y-value)

The function \[f(x) = \sin x\] has all real numbers in its domain, but its range is \[ - 1 \leqslant \sin x \leqslant 1\]. The values of the cosine function are different, depending on whether the angle is in degrees or radians.

Now we have been given the function, \[f(x) = {\sin ^2}x - 3\sin x + 2\].

Let us make the function in the form of \[{\left( {a - b} \right)^2}\]. To make it in this form add and subtract it with \[\dfrac{3}{2}\].

Thus simplify the expression that has been formed until we get a value for \[f(x)\].

\[
  f(x) = {\sin ^2}x - 3\sin x + 2 \ \\

  f(x) = {\left( {\sin x} \right)^2} - 2 \times \dfrac{3}{2} \times \sin x + {\left( {\dfrac{3}{2}} \right)^2} - {\left( {\dfrac{3}{2}} \right)^2} + 2 \ \\

  f(x) = {\left( {\sin x - \dfrac{3}{2}} \right)^2} - \dfrac{9}{4} + 2 \ \\

  f(x) = {\left( {\sin x - \dfrac{3}{2}} \right)^2} - \left( {\dfrac{{9 + 8}}{4}} \right) \ \\

  f(x) = {\left( {\sin x - \dfrac{3}{2}} \right)^2} - \dfrac{1}{4} \ \\
\]

Thus we got the function as,

\[f(x) = {\left( {\sin x - \dfrac{3}{2}} \right)^2} - \dfrac{1}{4}\]

We know that the range of \[\sin x\] is \[[ - 1,1]\]. Thus let us modify it to get the required range.

\[ - 1 \leqslant sinx \leqslant 1\]

Subtract \[\dfrac{3}{2}\]in the above expression and simplify it.

\[

   - 1 - \dfrac{3}{2} \leqslant \sin x - \dfrac{3}{2} \leqslant 1 - \dfrac{3}{2} \ \\

  \dfrac{{ - 2 - 3}}{2} \leqslant \sin x - \dfrac{3}{2} \leqslant \dfrac{{2 - 3}}{2} \ \\

   - \dfrac{5}{2} \leqslant \sin x - \dfrac{3}{2} \leqslant - \dfrac{1}{2} \ \\

\]

Now let us take the square of the expression and simplify it.

\[

  {\left( { - \dfrac{1}{2}} \right)^2} \leqslant {\left( {\sin x - \dfrac{3}{2}} \right)^2} \leqslant {\left( { - \dfrac{5}{2}} \right)^2} \ \\

  \dfrac{1}{4} \leqslant {\left( {\sin x - \dfrac{3}{2}} \right)^2} \leqslant \dfrac{{25}}{4} \ \\

  \dfrac{1}{4} - \dfrac{1}{4} \leqslant {\left( {\sin x - \dfrac{3}{2}} \right)^2} - \dfrac{1}{4} \leqslant \dfrac{{25}}{4} - \dfrac{1}{4} \ \\

  0 \leqslant {\left( {\sin x - \dfrac{3}{2}} \right)^2} - \dfrac{1}{4} \leqslant \dfrac{{24}}{4} \ \\

  0 \leqslant {\left( {\sin x - \dfrac{3}{2}} \right)^2} - \dfrac{1}{4} \leqslant 6 \ \\

\]

We got, \[f(x) = {\left( {\sin x - \dfrac{3}{2}} \right)^2} - \dfrac{1}{4}\], Thus replace it in the above expression.

\[0 \leqslant f(x) \leqslant 6\]

Thus from this we got the range of our function \[f(x)\]as,

\[f(x) = \left( {0,6} \right]\].

Hence we got the range of the function \[f(x) = {\sin ^2}x - 3\sin x + 2\] as \[\left( {0,6} \right]\].

Note: You can even substitute \[f(x)\]as \[y\], which is much easier to identify. By seeing the question, you should be able to identify which property you should be using in order to simplify it. It is important that you should know the range and domain of different functions such as \[\sin x,\cos x,\tan x\]etc, so that you might be able to find the range of functions related to it.