Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
To find the inverse function of [tex]\( g(x) = \frac{x-2}{5} \)[/tex], let's work through the steps systematically.
1. Write the function [tex]\( g(x) \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = \frac{x - 2}{5} \][/tex]
2. To find the inverse function, swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = \frac{y - 2}{5} \][/tex]
3. Solve the equation for [tex]\( y \)[/tex]:
- First, eliminate the fraction by multiplying both sides by 5:
[tex]\[ 5x = y - 2 \][/tex]
- Then, isolate [tex]\( y \)[/tex] by adding 2 to both sides of the equation:
[tex]\[ 5x + 2 = y \][/tex]
- Therefore, the inverse function is:
[tex]\[ y = 5x + 2 \][/tex]
So, the inverse of [tex]\( g(x) = \frac{x - 2}{5} \)[/tex] is [tex]\( g^{-1}(x) = 5x + 2 \)[/tex].
Now, considering the function [tex]\( f(x) = 5x + 2 \)[/tex], we observe that:
- [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are inverse functions of each other. This means that if you compose [tex]\( f \)[/tex] and [tex]\( g \)[/tex] (either [tex]\( f(g(x)) \)[/tex] or [tex]\( g(f(x)) \)[/tex]), you should get back [tex]\( x \)[/tex].
To verify:
- [tex]\( f(g(x)) = f\left(\frac{x-2}{5}\right) \)[/tex]:
[tex]\[ f\left(\frac{x-2}{5}\right) = 5 \left(\frac{x-2}{5}\right) + 2 = x - 2 + 2 = x \][/tex]
- [tex]\( g(f(x)) = g(5x + 2) \)[/tex]:
[tex]\[ g(5x + 2) = \frac{(5x + 2) - 2}{5} = \frac{5x}{5} = x \][/tex]
This confirms that:
[tex]\[ f(x) = 5x + 2 \quad \text{and} \quad g(x) = \frac{x-2}{5} \][/tex]
are indeed inverse functions of each other, ensuring that they have a unique and symmetric relationship as functions that essentially "undo" each other.
1. Write the function [tex]\( g(x) \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = \frac{x - 2}{5} \][/tex]
2. To find the inverse function, swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = \frac{y - 2}{5} \][/tex]
3. Solve the equation for [tex]\( y \)[/tex]:
- First, eliminate the fraction by multiplying both sides by 5:
[tex]\[ 5x = y - 2 \][/tex]
- Then, isolate [tex]\( y \)[/tex] by adding 2 to both sides of the equation:
[tex]\[ 5x + 2 = y \][/tex]
- Therefore, the inverse function is:
[tex]\[ y = 5x + 2 \][/tex]
So, the inverse of [tex]\( g(x) = \frac{x - 2}{5} \)[/tex] is [tex]\( g^{-1}(x) = 5x + 2 \)[/tex].
Now, considering the function [tex]\( f(x) = 5x + 2 \)[/tex], we observe that:
- [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are inverse functions of each other. This means that if you compose [tex]\( f \)[/tex] and [tex]\( g \)[/tex] (either [tex]\( f(g(x)) \)[/tex] or [tex]\( g(f(x)) \)[/tex]), you should get back [tex]\( x \)[/tex].
To verify:
- [tex]\( f(g(x)) = f\left(\frac{x-2}{5}\right) \)[/tex]:
[tex]\[ f\left(\frac{x-2}{5}\right) = 5 \left(\frac{x-2}{5}\right) + 2 = x - 2 + 2 = x \][/tex]
- [tex]\( g(f(x)) = g(5x + 2) \)[/tex]:
[tex]\[ g(5x + 2) = \frac{(5x + 2) - 2}{5} = \frac{5x}{5} = x \][/tex]
This confirms that:
[tex]\[ f(x) = 5x + 2 \quad \text{and} \quad g(x) = \frac{x-2}{5} \][/tex]
are indeed inverse functions of each other, ensuring that they have a unique and symmetric relationship as functions that essentially "undo" each other.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.