Question

The minimum value of the function \( f(x) = x^4 - 4x - 5 \), where \( x \in \mathbb{R} \), is:

• A. -7
• B. 7
• C. 8
• D. -8
• E. 0

Hint:

When analyzing the minimum or maximum of polynomial functions, checking the sign of the second derivative at critical points can help determine if they are minima, maxima, or saddle points.

Answer 1

The Correct Option is D

To find the minimum value of \( f(x) \), first compute the first derivative: \[ f'(x) = 4x^3 - 4 \] Set the derivative equal to zero to find critical points: \[ 4x^3 - 4 = 0 \] \[ x^3 = 1 \] \[ x = 1 \] Next, compute the second derivative to determine the nature of the critical point: \[ f''(x) = 12x^2 \] \[ f''(1) = 12(1)^2 = 12>0 \] Since \( f''(1)>0 \), the function has a local minimum at \( x = 1 \). Now, evaluate \( f(x) \) at \( x = 1 \): \[ f(1) = 1^4 - 4 \cdot 1 - 5 = 1 - 4 - 5 = -8 \] Given the fourth power of \( x \) in \( f(x) \), \( f(x) \to \infty \) as \( x \to \pm\infty \). Thus, the minimum value of \( f(x) \) on \( \mathbb{R} \) occurs at \( x = 1 \) and is: \[ -8 \]