To find the inverse of the function [tex]\( f(x) = \sqrt{11x + 12} \)[/tex], we need to follow a systematic approach to derive [tex]\( f^{-1}(x) \)[/tex].
1. Rewrite the function with [tex]\( y \)[/tex] instead of [tex]\( f(x) \)[/tex]:
[tex]\[ y = \sqrt{11x + 12} \][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = \sqrt{11y + 12} \][/tex]
3. Square both sides to eliminate the square root:
[tex]\[ x^2 = 11y + 12 \][/tex]
4. Solve for [tex]\( y \)[/tex]:
[tex]\[ 11y = x^2 - 12 \][/tex]
[tex]\[ y = \frac{x^2 - 12}{11} \][/tex]
Thus, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x^2 - 12}{11} \][/tex]
Next, we need to determine the domain of the inverse function. The domain of the inverse function corresponds to the range of the original function.
5. Determine the domain of the original function [tex]\( f(x) \)[/tex]:
The function [tex]\( f(x) = \sqrt{11x + 12} \)[/tex] requires the expression under the square root to be non-negative:
[tex]\[ 11x + 12 \geq 0 \][/tex]
[tex]\[ 11x \geq -12 \][/tex]
[tex]\[ x \geq \frac{-12}{11} \][/tex]
Since the domain of the original function is [tex]\( x \geq \frac{-12}{11} \)[/tex], this is the range of the inverse function.
Therefore, the correct inverse function and its appropriate domain from the options given are:
[tex]\[ f^{-1}(x) = \frac{x^2 - 12}{11}, \quad x \geq \frac{-12}{11} \][/tex]
However, looking closely at the provided options, it appears no option matches [tex]\( x \geq \frac{-12}{11} \)[/tex]. Thus, there might be an error in the options provided or a typo. Please verify the correct boundaries.
Since we need to choose from the given options:
[tex]\[ f^{-1}(x) = \frac{x^2 - 12}{11}, \quad x \geq 0 \][/tex]
is the most reasonable option given the typical range might start from zero for inverse practical considerations in many cases.
Therefore, the answer, while adjusted to the context of the given options, is:
[tex]\[ f^{-1}(x) = \frac{x^2 - 12}{11}, \quad x \geq 0 \][/tex]
