Answer :
To verify that [tex]\( g(x) = \frac{1}{5} x + 5 \)[/tex] is the inverse of [tex]\( f(x) = 5x - 25 \)[/tex], we need to show that [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex]. That is, applying one function after the other should yield the original input [tex]\( x \)[/tex].
Let's start by checking [tex]\( g(f(x)) \)[/tex]:
1. Compute [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 5x - 25 \][/tex]
2. Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex] to find [tex]\( g(f(x)) \)[/tex]:
[tex]\[ g(f(x)) = g(5x - 25) \][/tex]
[tex]\[ g(t) = \frac{1}{5} t + 5 \quad \text{(where \( t = 5x - 25 \))} \][/tex]
Thus:
[tex]\[ g(5x - 25) = \frac{1}{5} (5x - 25) + 5 \][/tex]
3. Simplify [tex]\( g(5x - 25) \)[/tex]:
[tex]\[ g(5x - 25) = \frac{1}{5} \cdot (5x - 25) + 5 \][/tex]
[tex]\[ g(5x - 25) = x - 5 + 5 \][/tex]
[tex]\[ g(5x - 25) = x \][/tex]
Since [tex]\( g(f(x)) = x \)[/tex], one direction of the inverse function relationship is verified.
Next, let's verify [tex]\( f(g(x)) \)[/tex]:
1. Compute [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = \frac{1}{5} x + 5 \][/tex]
2. Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex] to find [tex]\( f(g(x)) \)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{1}{5} x + 5\right) \][/tex]
[tex]\[ f(t) = 5t - 25 \quad \text{(where \( t = \frac{1}{5} x + 5 \))} \][/tex]
Thus:
[tex]\[ f\left(\frac{1}{5} x + 5\right) = 5\left(\frac{1}{5} x + 5\right) - 25 \][/tex]
3. Simplify [tex]\( f\left(\frac{1}{5} x + 5\right) \)[/tex]:
[tex]\[ f\left(\frac{1}{5} x + 5\right) = 5 \cdot \left(\frac{1}{5} x + 5\right) - 25 \][/tex]
[tex]\[ f\left(\frac{1}{5} x + 5\right) = x + 25 - 25 \][/tex]
[tex]\[ f\left(\frac{1}{5} x + 5\right) = x \][/tex]
Since [tex]\( f(g(x)) = x \)[/tex], the other direction of the inverse function relationship is verified.
Therefore, the expression that could be used to verify [tex]\( g(x) \)[/tex] is the inverse of [tex]\( f(x) \)[/tex] is:
[tex]\[ \frac{1}{5}(5x - 25) + 5 \][/tex]
as it simplifies to [tex]\( x \)[/tex], ensuring [tex]\( g(f(x)) = x \)[/tex].
The correct expression from the given options is:
[tex]\[ \frac{1}{5}(5x - 25) + 5 \][/tex]
Let's start by checking [tex]\( g(f(x)) \)[/tex]:
1. Compute [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 5x - 25 \][/tex]
2. Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex] to find [tex]\( g(f(x)) \)[/tex]:
[tex]\[ g(f(x)) = g(5x - 25) \][/tex]
[tex]\[ g(t) = \frac{1}{5} t + 5 \quad \text{(where \( t = 5x - 25 \))} \][/tex]
Thus:
[tex]\[ g(5x - 25) = \frac{1}{5} (5x - 25) + 5 \][/tex]
3. Simplify [tex]\( g(5x - 25) \)[/tex]:
[tex]\[ g(5x - 25) = \frac{1}{5} \cdot (5x - 25) + 5 \][/tex]
[tex]\[ g(5x - 25) = x - 5 + 5 \][/tex]
[tex]\[ g(5x - 25) = x \][/tex]
Since [tex]\( g(f(x)) = x \)[/tex], one direction of the inverse function relationship is verified.
Next, let's verify [tex]\( f(g(x)) \)[/tex]:
1. Compute [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = \frac{1}{5} x + 5 \][/tex]
2. Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex] to find [tex]\( f(g(x)) \)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{1}{5} x + 5\right) \][/tex]
[tex]\[ f(t) = 5t - 25 \quad \text{(where \( t = \frac{1}{5} x + 5 \))} \][/tex]
Thus:
[tex]\[ f\left(\frac{1}{5} x + 5\right) = 5\left(\frac{1}{5} x + 5\right) - 25 \][/tex]
3. Simplify [tex]\( f\left(\frac{1}{5} x + 5\right) \)[/tex]:
[tex]\[ f\left(\frac{1}{5} x + 5\right) = 5 \cdot \left(\frac{1}{5} x + 5\right) - 25 \][/tex]
[tex]\[ f\left(\frac{1}{5} x + 5\right) = x + 25 - 25 \][/tex]
[tex]\[ f\left(\frac{1}{5} x + 5\right) = x \][/tex]
Since [tex]\( f(g(x)) = x \)[/tex], the other direction of the inverse function relationship is verified.
Therefore, the expression that could be used to verify [tex]\( g(x) \)[/tex] is the inverse of [tex]\( f(x) \)[/tex] is:
[tex]\[ \frac{1}{5}(5x - 25) + 5 \][/tex]
as it simplifies to [tex]\( x \)[/tex], ensuring [tex]\( g(f(x)) = x \)[/tex].
The correct expression from the given options is:
[tex]\[ \frac{1}{5}(5x - 25) + 5 \][/tex]
