Question

The function $ y = x^3 - ax^2 + 48x + 7 $ is increasing for all real values of $ x $. Find the interval in which $ a $ lies:

• A. \( (-14, 14) \)
• B. \( (-12, 12) \)
• C. \( (-16, 16) \)
• D. \( (-21, 21) \)

Hint:

For a quadratic to always be positive, ensure the leading coefficient is positive and discriminant is negative.

Answer 1

The Correct Option is B

For the function to be increasing for all real \( x \), its first derivative must be positive for all \( x \): \[ f(x) = x^3 - ax^2 + 48x + 7 \Rightarrow f'(x) = 3x^2 - 2ax + 48 \] We want: \[ f'(x)>0 \quad \forall x \in \mathbb{R} \Rightarrow \text{Quadratic } 3x^2 - 2ax + 48>0 \text{ for all } x \] This happens when the discriminant \( D<0 \): \[ D = (-2a)^2 - 4 \cdot 3 \cdot 48 = 4a^2 - 576<0 \Rightarrow a^2<144 \Rightarrow a \in (-12, 12) \] Hence: \[ \boxed{a \in (-12, 12)} \]