Answer :
To determine whether the function [tex]\( f(x) = \sqrt{7x + 5} \)[/tex] has an inverse, we need to check if the function is one-to-one. A function is one-to-one if every value of [tex]\( f(x) \)[/tex] corresponds to exactly one value of [tex]\( x \)[/tex].
To check if [tex]\(f(x)\)[/tex] is one-to-one, we can use the horizontal line test: if any horizontal line intersects the graph of the function at most once, then the function is one-to-one.
For the function [tex]\( f(x) = \sqrt{7x + 5} \)[/tex]:
- The function involves a square root, which is typically one-to-one for the principal square root (which takes only the positive root).
- The expression inside the square root, [tex]\(7x + 5\)[/tex], is a linear function and hence strictly increasing.
- Hence, [tex]\( \sqrt{7x + 5} \)[/tex] is also strictly increasing because the composition of an increasing function with the square root (which is increasing for non-negative values) remains increasing.
Therefore, [tex]\( f(x) = \sqrt{7x + 5} \)[/tex] is a one-to-one function and thus has an inverse.
Next, we will find the inverse function [tex]\( f^{-1}(x) \)[/tex].
1. Start by writing the equation for [tex]\( f \)[/tex]:
[tex]\[ y = \sqrt{7x + 5} \][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse:
[tex]\[ x = \sqrt{7y + 5} \][/tex]
3. Solve for [tex]\( y \)[/tex]:
- Square both sides to eliminate the square root:
[tex]\[ x^2 = 7y + 5 \][/tex]
- Isolate [tex]\( y \)[/tex]:
[tex]\[ x^2 - 5 = 7y \][/tex]
[tex]\[ y = \frac{x^2 - 5}{7} \][/tex]
4. Thus, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x^2 - 5}{7} \][/tex]
Finally, since [tex]\( f(x) \)[/tex] involves the principal square root, which is defined for non-negative values of [tex]\( x \)[/tex], the domain of [tex]\( f^{-1}(x) \)[/tex] will be for [tex]\( x \geq 0 \)[/tex].
Thus, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x^2 - 5}{7}, \quad x \geq 0 \][/tex]
To check if [tex]\(f(x)\)[/tex] is one-to-one, we can use the horizontal line test: if any horizontal line intersects the graph of the function at most once, then the function is one-to-one.
For the function [tex]\( f(x) = \sqrt{7x + 5} \)[/tex]:
- The function involves a square root, which is typically one-to-one for the principal square root (which takes only the positive root).
- The expression inside the square root, [tex]\(7x + 5\)[/tex], is a linear function and hence strictly increasing.
- Hence, [tex]\( \sqrt{7x + 5} \)[/tex] is also strictly increasing because the composition of an increasing function with the square root (which is increasing for non-negative values) remains increasing.
Therefore, [tex]\( f(x) = \sqrt{7x + 5} \)[/tex] is a one-to-one function and thus has an inverse.
Next, we will find the inverse function [tex]\( f^{-1}(x) \)[/tex].
1. Start by writing the equation for [tex]\( f \)[/tex]:
[tex]\[ y = \sqrt{7x + 5} \][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse:
[tex]\[ x = \sqrt{7y + 5} \][/tex]
3. Solve for [tex]\( y \)[/tex]:
- Square both sides to eliminate the square root:
[tex]\[ x^2 = 7y + 5 \][/tex]
- Isolate [tex]\( y \)[/tex]:
[tex]\[ x^2 - 5 = 7y \][/tex]
[tex]\[ y = \frac{x^2 - 5}{7} \][/tex]
4. Thus, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x^2 - 5}{7} \][/tex]
Finally, since [tex]\( f(x) \)[/tex] involves the principal square root, which is defined for non-negative values of [tex]\( x \)[/tex], the domain of [tex]\( f^{-1}(x) \)[/tex] will be for [tex]\( x \geq 0 \)[/tex].
Thus, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x^2 - 5}{7}, \quad x \geq 0 \][/tex]
