If `f (x)= (px)/(e ^(x)) - (x ^(2))/(2) + x ` is a decreasing function for every `x le 0.` Find the least value of `p ^(2).`

Answer

To solve the problem, we need to determine the least value of \( p^2 \) such that the function

\[ f(x) = \frac{px}{e^x} - \frac{x^2}{2} + x \]

is a decreasing function for every \( x \leq 0 \).

### Step 1: Differentiate the function \( f(x) \)

To find when \( f(x) \) is decreasing, we need to find the derivative \( f'(x) \) and set it less than or equal to zero.

Using the quotient rule for the first term, we have:

\[ f'(x) = \frac{d}{dx}\left(\frac{px}{e^x}\right) - \frac{d}{dx}\left(\frac{x^2}{2}\right) + \frac{d}{dx}(x) \]

Calculating the derivative of the first term using the quotient rule:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = px \) and \( v = e^x \).

Thus,

\[ f'(x) = \frac{e^x \cdot p - px \cdot e^x}{(e^x)^2} - x + 1 \]

This simplifies to:

\[ f'(x) = \frac{p(e^x - x e^x)}{e^{2x}} - x + 1 \]

### Step 2: Simplify \( f'(x) \)

Factoring out \( e^x \):

\[ f'(x) = \frac{p e^x (1 - x)}{e^{2x}} - x + 1 = \frac{p(1 - x)}{e^x} - x + 1 \]

### Step 3: Set \( f'(x) \leq 0 \)

For \( f(x) \) to be decreasing for all \( x \leq 0 \):

\[ \frac{p(1 - x)}{e^x} - x + 1 \leq 0 \]

### Step 4: Analyze the inequality

At \( x = 0 \):

\[ \frac{p(1 - 0)}{e^0} - 0 + 1 \leq 0 \implies p + 1 \leq 0 \implies p \leq -1 \]

### Step 5: Find the least value of \( p^2 \)

The least value of \( p \) that satisfies this condition is \( p = -1 \). Therefore, we calculate \( p^2 \):

\[ p^2 = (-1)^2 = 1 \]

### Conclusion

The least value of \( p^2 \) is

\[ \boxed{1} \]