Question

Find the domain of the real valued function \[f(x) = \sqrt {2 - x} + \sqrt {1 + x} \]

Answer 1

Hint: We write the definition of domain of a function and the definition of a real valued function. Find the domain of the given real valued functions by equating the terms in the function to greater than or equal to 0.
* A function is called a real valued function if that function assigns a real number to each member of its domain i.e. the values of real valued functions are always real numbers (not complex).
* Domain of a function contains all those values that can be taken as an input such that we obtain the value of the function.

Complete step-by-step solution:
Here we are given real valued function \[f(x) = \sqrt {2 - x} + \sqrt {1 + x} \]
Since the function is real valued then all the values of the variable ‘x’ are real numbers.
We know for a function to be defined the terms in RHS need to be greater than or equal to 0
We write \[\sqrt {2 - x} \geqslant 0\] and \[\sqrt {1 + x} \geqslant 0\]
Square both sides of the equations
\[ \Rightarrow {\left( {\sqrt {2 - x} } \right)^2} \geqslant {0^2}\] and \[{\left( {\sqrt {1 + x} } \right)^2} \geqslant {0^2}\]
Cancel square root by square power on both sides of the equation
\[ \Rightarrow 2 - x \geqslant 0\] and \[1 + x \geqslant 0\]
Now shift all the constant terms in left hand side of the equation to right hand side of the equation
\[ \Rightarrow - x \geqslant - 2\] and \[x \geqslant - 1\]
Multiply both sides of the equation by -1 in first equation
\[ \Rightarrow x \leqslant 2\] and \[x \geqslant - 1\]
Write the value of x in form of intervals i.e. \[ - 1 \leqslant x \leqslant 2\]

So we can write the domain of the real-valued function is \[\left[ { - 1,2} \right]\]

Note: Many students make the mistake of not changing the sign of inequality when they multiply the inequality by negative sign, keep in mind the inequality sign becomes exactly opposite when we multiply negative numbers to both sides of the equation.