Question-21

Find the domain and range of \(\displaystyle f( x) =\sqrt{5-4x-x^{2}}\).

• Domain is \([0, \infty)\)
• Domain is \((-\infty, -5] \cup [1, \infty)\)
• Domain is \([-5, 1]\)
• Range is \([0, \infty)\)
• Range is \([0, 3]\)

Hint

Find the roots of the quadratic equation and the maximum value of the quadratic expression.

Answer

• Domain is \([0, \infty)\)
• Domain is \((-\infty, -5] \cup [1, \infty)\)
• Domain is \([-5, 1]\)
• Range is \([0, \infty)\)
• Range is \([0, 3]\)

Solution

For the domain, we have:

\[\begin{equation*} \begin{aligned} 5-4x-x^{2} & \geqslant 0\\ x^{2} +4x-5 & \leqslant 0\\ ( x+5)( x-1) & \leqslant 0\\ \Longrightarrow x & \in [ -5,1] \end{aligned} \end{equation*}\]

For the range, we see that the curve \(\displaystyle 5-4x-x^{2}\) is a concave downward parabola that has a maximum. So the range of \(\displaystyle f\) will be \(\displaystyle [ 0,\max( f( x))]\). To compute the maximum, we can compute the derivative, set it to zero and find the point of maxima. Alternatively, if the parabola is \(\displaystyle ax^{2} +bx+c\), the maximum occurs at \(\displaystyle \frac{-b}{2a}\). In this case, it occurs at \(\displaystyle -2\) and the corresponding function-value is \(\displaystyle 3\). Therefore, the range of \(\displaystyle f\) is \(\displaystyle [ 0,3]\).