Answer :
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.
