14. Defermine the vatuelgs of \( p \) for which ithe squation \( (1-p) x^{3}+x+(1-p)=0 \) wil have acpual teots.

To determine the values of \( p \) for which the equation \[ (1-p)x^3 + x + (1-p) = 0 \] will have actual roots, we need to analyze the conditions under which this cubic equation has real roots. ### Step 1: Identify the coefficients The equation can be rewritten in standard form as: \[ a x^3 + b x + c = 0 \] where: - \( a = 1 - p \) - \( b = 1 \) - \( c = 1 - p \) ### Step 2: Determine the conditions for real roots A cubic equation can have either one real root or three real roots. The discriminant of a cubic equation can help us determine the nature of the roots. However, for cubic equations, we can also analyze the behavior of the function. ### Step 3: Analyze the function The function can be expressed as: \[ f(x) = (1-p)x^3 + x + (1-p) \] To find the conditions for real roots, we can check the derivative to find critical points: \[ f'(x) = 3(1-p)x^2 + 1 \] ### Step 4: Analyze the derivative The critical points occur when \( f'(x) = 0 \): \[ 3(1-p)x^2 + 1 = 0 \] This gives us: \[ x^2 = -\frac{1}{3(1-p)} \] For \( x^2 \) to be non-negative, we need: \[ 1 - p > 0 \implies p < 1 \] ### Step 5: Check the behavior of the function - If \( p < 1 \), the function \( f(x) \) is a cubic function that can have either one or three real roots depending on the value of \( p \). - If \( p = 1 \), the equation simplifies to \( x + 0 = 0 \), which has one real root \( x = 0 \). - If \( p > 1 \), the leading coefficient \( 1 - p \) becomes negative, and the function will have one real root. ### Conclusion The equation will have actual roots for all values of \( p \). Thus, the values of \( p \) for which the equation has actual roots are: \[ p \in \mathbb{R} \] This means that for any real number \( p \), the equation will have at least one real root.