Answer :
To find the inverse function of [tex]\( f(x) = \sqrt{3x + 6} \)[/tex], follow these steps:
1. Rewrite the function using [tex]\( y \)[/tex]:
[tex]\[ y = \sqrt{3x + 6} \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
First, square both sides to get rid of the square root:
[tex]\[ y^2 = 3x + 6 \][/tex]
Next, isolate [tex]\( x \)[/tex] by first subtracting 6 from both sides:
[tex]\[ y^2 - 6 = 3x \][/tex]
Then, divide both sides by 3 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{y^2 - 6}{3} \][/tex]
3. Express the inverse function:
The inverse function [tex]\( f^{-1}(x) \)[/tex] is given by:
[tex]\[ f^{-1}(x) = \frac{x^2 - 6}{3} \][/tex]
Now, we need to determine the valid domain for the inverse. The function [tex]\( f(x) = \sqrt{3x + 6} \)[/tex] is defined for values of [tex]\( x \)[/tex] that make the expression under the square root non-negative. This translates to:
[tex]\[ 3x + 6 \ge 0 \][/tex]
[tex]\[ 3x \ge -6 \][/tex]
[tex]\[ x \ge -2 \][/tex]
However, since the original function [tex]\( f(x) \)[/tex] involves a square root and outputs non-negative numbers, the inverse function [tex]\( f^{-1}(x) \)[/tex] should be real and non-negative for [tex]\( x \geq 0 \)[/tex]:
Thus, the correct inverse and its domain are:
[tex]\[ f^{-1}(x) = \frac{x^2 - 6}{3}, \quad x \geq 0 \][/tex]
From the given choices, it matches:
[tex]\[ f^{-1}(x) = \frac{x^2 - 6}{3}, x \geq 0 \][/tex]
So, the correct option is:
1. [tex]\( f^{-1}(x) = \frac{x^2 - 6}{3}, x \geq 0 \)[/tex]
1. Rewrite the function using [tex]\( y \)[/tex]:
[tex]\[ y = \sqrt{3x + 6} \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
First, square both sides to get rid of the square root:
[tex]\[ y^2 = 3x + 6 \][/tex]
Next, isolate [tex]\( x \)[/tex] by first subtracting 6 from both sides:
[tex]\[ y^2 - 6 = 3x \][/tex]
Then, divide both sides by 3 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{y^2 - 6}{3} \][/tex]
3. Express the inverse function:
The inverse function [tex]\( f^{-1}(x) \)[/tex] is given by:
[tex]\[ f^{-1}(x) = \frac{x^2 - 6}{3} \][/tex]
Now, we need to determine the valid domain for the inverse. The function [tex]\( f(x) = \sqrt{3x + 6} \)[/tex] is defined for values of [tex]\( x \)[/tex] that make the expression under the square root non-negative. This translates to:
[tex]\[ 3x + 6 \ge 0 \][/tex]
[tex]\[ 3x \ge -6 \][/tex]
[tex]\[ x \ge -2 \][/tex]
However, since the original function [tex]\( f(x) \)[/tex] involves a square root and outputs non-negative numbers, the inverse function [tex]\( f^{-1}(x) \)[/tex] should be real and non-negative for [tex]\( x \geq 0 \)[/tex]:
Thus, the correct inverse and its domain are:
[tex]\[ f^{-1}(x) = \frac{x^2 - 6}{3}, \quad x \geq 0 \][/tex]
From the given choices, it matches:
[tex]\[ f^{-1}(x) = \frac{x^2 - 6}{3}, x \geq 0 \][/tex]
So, the correct option is:
1. [tex]\( f^{-1}(x) = \frac{x^2 - 6}{3}, x \geq 0 \)[/tex]
