Question

Find the range of \[f(x) = \sqrt {{{\sin }^2}x - 6\sin x + 9} + 3\].

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, \[f(x) = \sqrt {{{\sin }^2}x - 6\sin x + 9} + 3\].

Let us make the function in the form of \[{\left( {a - b} \right)^2}\]. We know its formula as,

\[
  f(x) = \sqrt {{{\sin }^2}x - 6\sin x + 9} + 3 \ \\
  {\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2} \ \\
  a = \sin x,b = 3 \ \\
  {\sin ^2}x - 6\sin x + 9 = {\left( {\sin x - 3} \right)^2} \ \\
  f(x) = \sqrt {{{\left( {\sin x - 3} \right)}^2}} + 3 \ \\
  f(x) = \left| {\sin x - 3} \right| + 3 \ \\
\]

Thus we got the function as, \[f(x) = \left| {\sin x - 3} \right| + 3\].
We know that the range of \[\sin x\] is \[[ - 1,1]\]. Thus let us modify it to get the required range.

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

Let us subtract 3 from the above expression and simplify it to get,

\[

   - 1 \leqslant \sin x \leqslant 1 \ \\

   - 1 - 3 \leqslant \sin x - 3 \leqslant 1 - 3 \ \\

   - 4 \leqslant \sin x - 3 \leqslant - 2 \ \\

 \]

Now take modulus of the above expression.

\[

  \left| { - 4} \right| \leqslant \left| {\sin x - 3} \right| \leqslant \left| { - 2} \right| \ \\

  4 \leqslant \left| {\sin x - 3} \right| \leqslant 2 \ \\

 \]

Now let us add 3 to the above expression to make it similar to \[f(x)\] we got.

\[

  2 \leqslant \left| {\sin x - 3} \right| \leqslant 4 \ \\

  2 + 3 \leqslant \left| {\sin x - 3} \right| + 3 \leqslant 4 + 3 \ \\

  5 \leqslant \left| {\sin x - 3} \right| + 3 \leqslant 7 \ \\

 \]

We earlier got that \[f(x) = \left| {\sin x - 3} \right| + 3\], thus replace it.

\[
  5 \leqslant \left| {\sin x - 3} \right| + 3 \leqslant 7 \ \\
  5 \leqslant f(x) \leqslant 7 \ \\
 \]

Thus from this we got the range of our function \[f(x)\] as,
\[f(x) = [5,7]\].

Hence we got the range of the function \[f(x) = \sqrt {{{\sin }^2}x - 6\sin x + 9} + 3\] as \[[5,7]\].

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.