The inverse relationship for the square function y = x2 with its entire domain of all real numbers is y = $\pm$ $\sqrt{x}$ which is not a function. An inverse function can be defined by restricting the domain of the square function. Thus the square root function g (x) = $\sqrt{x}$ is the inverse for f(x) = x2 where x is a non negative number.

Same way restricting the domain as x $\leq$ 0, the negative of g(x), h(x) = - $\sqrt{x}$ becomes the inverse for the square function. 

A square root function is an algebraic function as it is expressed as the square root of the variable. The parent square root function is given as
f(x) = $\sqrt{x}$
The graph of square root function is given below:

Square Root Function

The characteristics of square root function can be stated as:

  • The domain of the function is the set of all non negative real numbers.
  • The range of the function is the set of all non negative real numbers.
  • The graph meets both the x and y axis at (0, 0)
  • The function is increasing or the graph is rising up in the entire domain of the function.

The transformed equations of square root function are given below detailing the transformation involved.

Equation
Transformation involved
Example
g(x) = a$\sqrt{x}$ Vertical stretch by a factor of a g(x) = 3$\sqrt{x}$
g(x) = $\sqrt{b}$x Vertical stretch by a factor of 1/b g(x) = $\sqrt{2}$x
g(x) = -$\sqrt{x}$Reflection over x - axis.
Vertical reflection.
g(x) = $\sqrt{-x}$Reflection over y - axis.
Horizontal reflection
g(x) = $\sqrt{x}$ + k Vertical shift up by k units. When k is
negative shift down by k units.
g(x) = $\sqrt{x}$ + 4
g(x) = $\sqrt{x}$ - 5
g(x) = $\sqrt{x-h}$ Horizontal shift to right by h units.
g(x) = $\sqrt{x-3}$
g(x) = $\sqrt{x+h}$
Horizontal shift to left by h units.
g(x) = $\sqrt{x+4}$

Examples for a composite transformation:

g(x) = $-3\sqrt{x-2}+5$
The transformations occur in the following order.
  1. Reflection over x axis.
  2. Vertical Stretch by factor 3
  3. Horizontal shift to right by 2 units and vertical shift up by 5 units.

h(x) = -$\sqrt{2x +6}$ - 4
The equation has to be first rewritten as = -$\sqrt{2(x+3)}$ - 4. The transformations occur in the following order.

  1. Reflection over x axis.
  2. Horizontal stretch by factor $\frac{1}{2}$.
  3. Horizontal shift to left by 3 units and vertical shift down by 4 units.
We have already seen that the domain of square root function f(x) = $\sqrt{x}$ is the set of all non negative numbers, that is x $\geq$ 0.

For other functions involving square root, the domain is determined by setting similar condition on the expression inside the radical and solving the inequality arising out of the condition.

Solved Example

Question: Find the domain of f(x) = $2\sqrt{-3x+9}$
Solution:
 
For the function f(x) to be defined, the expression under the radical should be non negative.
-3x + 9 ≥ 0
     -3x ≥ -9
       x ≤ 3                 The domain of the function.
The domain of f(x) can be written in interval form as (-$\infty$, 3]
 

The range of the parent square root function f(x) = $\sqrt{x}$ is set of all non negative real numbers, i.e f(x) $\geq$ 0.

The vertical reflection and shift modifies the range of other functions involving square root.

Example:
f(x) = -$\sqrt{x+5}$ + 4
Here the expression $\sqrt{x+5}$ is always greater than equal to zero. This value is subtracted from 4 to get the function value. Hence 4 is the upper boundary for f(x).
Hence the range of f(x) is f(x) $\leq$ 4 or in interval notation(-$\infty$, 4].

g(x) = $\sqrt{x}$ - 3, In this case the least value of the function is -3, as a non negative number is added to -3 to get the function value.
The range of g(x) is hence g(x) $\geq$ -3 or [-3, $\infty$).
Let us write one function equation, when the transformations are given.

Solved Example

Question: Write the equation of g(x) which is got by applying the following transformations to square root function f(x) = $\sqrt{x}$.
  1. Reflection over x axis.
  2. Horizontal stretch of factor 2.
  3. Shift to right by 3 units and shift up by 5 units.

Solution:
 

Reflection over x axis will give the transformation g(x) = -√x

The horizontal stretch will make the equation as g(x) = -$\sqrt{\frac{1}{2}(x)}$

Shift to right by 3 units will alter the equation as g(x) = -$\sqrt{\frac{1}{2}(x-3)}$

The final equation applying the vertical shift of 5 units is
g(x) = -$\sqrt{\frac{1}{2}(x-3)}$ + 5
 

We have seen the graph of the function f(x) = $\sqrt{x}$. Some of the points on the graph are (0, 0), (1, 1) and (4, 2). The points for the graph of a transformed square root function can be got by applying the transformation rule to the points on the parent graph.

Example:
Graph g(x) = -$\sqrt{2x+5}$ + 3 applying the transformation rule to points on the parent graph f(x) = $\sqrt{x}$.

g(x) can be rewritten to separate the horizontal stretch and shift as follows

g(x) = -$\sqrt{2(x+2.5)}$ + 3

So the transformations to be considered are reflection over x axis, horizontal stretch by factor $\frac{1}{2}$, horizontal shift to the left by 2.5 units and vertical shift up by 3 units.

The Rule to get the ordered pairs of g(x) applying transformation on the ordered pairs of the parent graph f(x) = $\sqrt{x}$ is

 f(x) 
   →              g(x)
     
(x, y)
   → ($\frac{x}{2}$ - 2.5, -y + 3)

Verbally the above rule can be stated as:
Divide the x coordinate of a point of f(x) and subtract 2.5 to get the x coordinate of the corresponding point on g(x)

Change the sign of y coordinate of a point of f(x) and add 3 to it to get the y coordinate of the corresponding point on g(x).

(x, f(x))
 (x, g(x))
(0, 0)
 (-2.5, 3)
 (1, 1)  (-2, 2)
 (4, 2)  (-0.5, 1)
  (9,3) (2, 0)

Plot the ordered pairs found for g(x) and join them smoothly to get the graph.

The graph of g(x) along with the parent f(x) (shown in dotted lines) is shown below:

Graphing Square Root Functions

The derivative of the parent function y = $\sqrt{x}$ can be found by replacing the radical with index $\frac{1}{2}$ and applying the power formula for derivatives.

y = $\sqrt{x}$ = x1/2.

$\frac{dy}{dx}$ = $\frac{1}{2}$ $x^{\frac{1}{2}-1}$

= $\frac{1}{2}$ $x^{-\frac{1}{2}}$

= $\frac{1}{2}$ $\frac{1}{\sqrt{x}}$

The derivative of a transformed square root function can be found in a similar manner applying chain rule.

Solved Example

Question: g(x) = $\sqrt{2x-5}$ = $(2x-5)^{\frac{1}{2}}$
Solution:
 
g '(x) = $\frac{1}{2}$ $(2x-5)^{\frac{1}{2}-1}$.$\frac{d}{dx}$$(2x-5)$

        = $\frac{1}{2}$ $(2x-5)^{-\frac{1}{2}}$.2

        = $(2x-5)^{-\frac{1}{2}}$

        = $\frac{1}{\sqrt{2x-5}}$
 


Solved Examples

Question 1: Find the domain and range of the function f(x) = 2$\sqrt{2x-8}$ + 5
Solution:
 
f(x) is defined when the expression under the radical sign is non negative.
   2x - 8 $\geq$ 0     ⇒     x $\geq$ 4
   Hence the domain in interval notation [4, $\infty$)

   f(x) assumes 5 as the smallest value when the 2x - 8 = 0.  Hence the range of the function is f(x) $\geq$ 5 or [5, $\infty$).
 

Question 2: Find the transformation rule to be used for graphing h(x) = $2\sqrt{-5x+5}$ - 3.
Solution:
 
The transformations can be listed rewriting the equation as h(x) = $2\sqrt{-5(x-1)}$ - 3
  1. Vertical stretch by a scale factor 2.
  2. Horizontal reflection (Reflection over y- axis)
  3. Horizontal compress by a factor $\frac{1}{5}$
  4. Horizontal shift to right by one unit and vertical shift down by 3 units

Hence the transformation rule can be written as

 f(x) = x 
   →      h(x) =$2\sqrt{-5x+5}$ - 3
     
(x, y)
   → (- $\frac{x}{5}$ + 1, 2y - 3)