Question

How do you find the domain and range of $ \sqrt {8 - x} $ ?

Answer 1

Hint: As we can see that there is a root in the equation, this means that the value of the function can never be negative as the root of a negative value is imaginary and not real.

Complete step by step solution:
The value under the root should always be greater or equal to zero for a positive and real outcome, so we can say that the value of $ \sqrt {8 - x} $ will be always positive or equal to zero and for $ \sqrt {8 - x} $ to be positive the value of $ 8 - x $ should be equal to or greater than $ 0 $ , so the domain can be found as,
Domain, Df:
 $ \Rightarrow 8 - x \geqslant 0 $
Multiplying the equation with minus sign will give us,
 $ \Rightarrow x - 8 \leqslant 0 $
Moving $ 8 $ to the other side of equality sign will give us the following equation,
 $ \Rightarrow x \leqslant 8 $
This means that the value of x can be taken from $ - \infty $ to $ 8 $ in order to get a real value from the equation, hence the domain will be denoted as,
Df: $ \{ - \infty < x \leqslant 8\} $
Now for range,
Firstly let the equation be denoted by ‘y’, so it will be written as,
 $ \Rightarrow y = \sqrt {8 - x} $
Now the possible value of y will be called as range of the function and as we have discussed above also that the value of the root to be real should always be positive, hence tha range of the function can be written as,
\[\Rightarrow {R_f} = \{ 0 \leqslant y < \infty \} \]

Note: When we multiply an equation having greater than or less than sign with minus sign than the greater than or less than sign gets reversed as seen above also. This is to be noted as a rule of mathematics. Also the value under the root can never be negative for a real outcome.