Graphical Analysis of Range of Square Root Functions
The range of a function \( y = f(x) \) is the set of values \( y \) takes for all values of \( x \) within the domain of \( f \).
What is the range of \( f(x) = \sqrt{x} \)?
The domain of \( f \) above is the set of all values of \( x \) in the interval \([0, +\infty)\).
As \( x \) takes values from \( 0 \) to \( +\infty \), \( \sqrt{x} \) takes values from \( \sqrt{0} \) to \( \sqrt{\infty} \). Hence, the range of \( \sqrt{x} \) is given by the interval \([0, +\infty)\).
The range of functions of the form \( \sqrt{x + k} \) is also given by the interval \([0, +\infty)\), as horizontal shifts do not affect the range. We can express this in inequality form as:
\( y \geq 0 \quad \text{or} \quad \sqrt{x + k} \geq 0 \)
Examples with Solutions
Example 1
Find the range of function \( f \) defined by:
\[ f(x) = \sqrt{x - 1} \]View Solution
We know that the range of the basic function \( f(x) = \sqrt{x} \) is given by the interval \([0, +\infty)\).
The graph of the given function \( f(x) = \sqrt{x - 1} \) is the graph of \( \sqrt{x} \) shifted 1 unit to the right. A horizontal shift does not affect the range. Hence, the range of \( f(x) = \sqrt{x - 1} \) is also given by the interval:
\[ [0, +\infty) \]Example 2
Find the range of function \( f \) defined by:
\[ f(x) = -\sqrt{x + 2} \]View Solution
Start with the range of the expression \( \sqrt{x + 2} \) in inequality form:
\[ \sqrt{x + 2} \geq 0 \]Multiply both sides of the inequality by -1 to obtain:
\[ -\sqrt{x + 2} \leq 0 \]The range of the expression \( -\sqrt{x + 2} \), which is also the range of the given function, is given by the interval:
\[ (-\infty, 0] \]Example 3
Find the range of function \( f \) defined by:
\[ f(x) = -2\sqrt{x + 3} + 5 \]View Solution
The range of values of \( \sqrt{x + 3} \) can be written as an inequality:
\[ \sqrt{x + 3} \geq 0 \]Multiply both sides by -2 and reverse the inequality sign:
\[ -2\sqrt{x + 3} \leq 0 \]Add 5 to both sides of the inequality:
\[ -2\sqrt{x + 3} + 5 \leq 5 \]The range of values of the expression on the left side, which is the range of the given function, is given by the interval:
\[ (-\infty, 5] \]Example 4
Find the range of function \( f \) defined by:
\[ f(x) = \sqrt{16 - x^2} \]View Solution
First, find the domain of the function by setting the radicand non-negative:
\[ 16 - x^2 \geq 0 \]The solution set to this inequality is the domain of \( f(x) \), given by the interval:
\[ [-4, 4] \]The range of values of \( 16 - x^2 \) for \( x \) in \( [-4, 4] \) is given by the interval \([0, 16]\) since the graph is a parabola with a maximum at \((0, 16)\).
The given function is the square root of \( 16 - x^2 \), and therefore has the range:
\[ [\sqrt{0}, \sqrt{16}] = [0, 4] \]
Example 5
Find the range of function \( f \) defined by:
\[ f(x) = \sqrt{x^2 - 25} \]View Solution
The domain of the given function requires:
\[ x^2 - 25 \geq 0 \]The solution set is given by the interval:
\[ (-\infty, -5] \cup [5, +\infty) \]For \( x \) in this domain, the range of the expression \( x^2 - 25 \) is \([0, +\infty)\). Taking the square root gives the range:
\[ [\sqrt{0}, \sqrt{+\infty}) = [0, +\infty) \]
Example 6
Find the range of function \( f \) defined by:
\[ f(x) = \sqrt{x^2 - 4x + 8} \]View Solution
The domain of the function requires:
\[ x^2 - 4x + 8 \geq 0 \]The discriminant of \( x^2 - 4x + 8 \) is \( (-4)^2 - 4(1)(8) = -16 \). Since the discriminant is negative and the leading coefficient is positive, the quadratic expression is always positive for all real numbers (the domain is all real numbers).
Rewriting the expression in vertex form:
\[ x^2 - 4x + 8 = (x - 2)^2 + 4 \]The graph of \( (x - 2)^2 + 4 \) is a parabola with a minimum at vertex \((2, 4)\). Hence, the range of \( x^2 - 4x + 8 \) is \([4, +\infty)\).
Taking the square root gives the range:
\[ [\sqrt{4}, \sqrt{+\infty}) = [2, +\infty) \]