If g(x) is monotonically increasing and f(x) is monotonically decreasing for x R and if (gof) (x) is defined for x e R, then prove that (gof)(x) will be monotonically decreasing function. Hence prove that` (gof) (x +1)leq (gof) (x-1). `

Options

A. `g(f(ax+1))gtg(f(ax-1))` if `alt0`
B. `g(g(ax+1)gtg(g(ax-1))` if `alt0`
C. `g(f(ax+1)ltg(f(ax-1))` if `agt0`
D. `g(g(ax+1))gtg(g(ax-1))` if `agt0`

Explanation

To solve the problem, we need to prove that the composition of the functions \( g(f(x)) \) is monotonically decreasing given that \( g(x) \) is monotonically increasing and \( f(x) \) is monotonically decreasing. Then, we will show that \( g(f(x + 1)) \leq g(f(x - 1)) \).

### Step-by-Step Solution:

1. **Understanding Monotonicity**:
- A function \( g(x) \) is monotonically increasing if for any \( x_1 < x_2 \), \( g(x_1) \leq g(x_2) \).
- A function \( f(x) \) is monotonically decreasing if for any \( x_1 < x_2 \), \( f(x_1) \geq f(x_2) \).

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