Question

Find the domain and the range of following functions:
(i) $f\left( x \right) = {x^2}$
(ii) $f\left( x \right) = \sqrt {\left( {x - 1} \right)\left( {3 - x} \right)} $

Answer 1

Hint: In the above question, is based on the concept of finding out the range and domain of the given function. The main approach towards solving this kind of question is to find out the set of the values where the function is defined and also the set of values that function can take i.e., the range.

Complete step-by-step answer:
A set of values defined for a function is called domain and the range is the function that can take these values. Domain is also called a replacement set and range is called a solution set.
So, in the given question above we will first find the domain and range of the function,
$f\left( x \right) = {x^2}$
So the graph of the \[{x^2}\] will look like a parabola with the vertex at the origin.
So we need to find out all possible outcomes, which depends on the input.On giving input to the value of x we will always get a positive real number.So the domain here is all real numbers.
\[Domain = R\]
For range,since the parabola is on the positive y axis,so we get positive real values where \[f(x) \geqslant 0\].
\[Range = \left\{ {f\left( x \right) \in R|f\left( x \right) \geqslant 0} \right\}\]
Now finding domain for $f\left( x \right) = \sqrt {\left( {x - 1} \right)\left( {3 - x} \right)} $ we get,
\[
  Domain:\left( {x - 1} \right)\left( {3 - x} \right) \geqslant 0 \Rightarrow \left( {x - 1} \right)\left( {3 - x} \right) \leqslant 0 \\
  x \in \left[ {1,3} \right] \\
 \]
\[
  Range: \\
  y = \sqrt {\left( {x - 1} \right)\left( {3 - x} \right)} \\
   \Rightarrow {y^2} = 3x - {x^2} - 3 + x \\
   \Rightarrow {y^2} = - {x^2} + 4x - 3 \\
   \Rightarrow x = \dfrac{{ - 4 \pm \sqrt {{{16}^2} - 4\left( {3 + {y^2}} \right)} }}{2} \\
 \]
The above equation is defined for
\[
  16 - 4\left( {3 + {y^2}} \right) = 0 \\
   \Rightarrow 4 \geqslant \left( {3 + {y^2}} \right) \\
   \Rightarrow 1 \geqslant {y^2} \geqslant 0 \\
   \Rightarrow 1 - {y^2} \geqslant 0 \\
   \Rightarrow \left( {1 - y} \right)\left( {1 + y} \right) \geqslant 0 \\
  Range:y \in ( - \infty ,1] \cup [1,\infty ) \\
 \]

Note: An important thing to note is that the type of parenthesis used in the range has a significance. The square brackets indicate that the endpoints mentioned are included in the range and the round brackets indicate that the endpoints are not included in the range.