If `f(x)=1+2x and g(x) = x/2` , then: `(f@g)(x)-(g@f)(x)=`

Options

A. 4
B. `1/4`
C. 2
D. `1/2`

Answer

To solve the problem, we need to find the expression \((f@g)(x) - (g@f)(x)\) where \(f(x) = 1 + 2x\) and \(g(x) = \frac{x}{2}\).

### Step-by-Step Solution:

1. **Find \(g(x)\)**:
\[ g(x) = \frac{x}{2} \]

2. **Find \(f(g(x))\)**:
We need to substitute \(g(x)\) into \(f(x)\):
\[ f(g(x)) = f\left(\frac{x}{2}\right) = 1 + 2\left(\frac{x}{2}\right) \]
Simplifying this:
\[ f(g(x)) = 1 + x = x + 1 \]

3. **Find \(f(x)\)**:
\[ f(x) = 1 + 2x \]

4. **Find \(g(f(x))\)**:
Now we substitute \(f(x)\) into \(g(x)\):
\[ g(f(x)) = g(1 + 2x) = \frac{1 + 2x}{2} \]
Simplifying this:
\[ g(f(x)) = \frac{1}{2} + x \]

5. **Now calculate \(f(g(x)) - g(f(x))\)**:
We have:
\[ f(g(x)) = x + 1 \]
\[ g(f(x)) = \frac{1}{2} + x \]
Therefore:
\[ f(g(x)) - g(f(x)) = (x + 1) - \left(\frac{1}{2} + x\right) \]
Simplifying this:
\[ = x + 1 - \frac{1}{2} - x \]
The \(x\) terms cancel out:
\[ = 1 - \frac{1}{2} = \frac{1}{2} \]

### Final Answer:
\[ (f@g)(x) - (g@f)(x) = \frac{1}{2} \]