Question

What is the nature of the function $ f(x) = x^3 - 3x^2 + 4x $ on real numbers?

• A. Strictly decreasing
• B. Decreasing
• C. Increasing
• D. Constant

Hint:

To determine the nature of a function, check the sign of its first derivative. If the first derivative is always positive, the function is increasing. If it is always negative, the function is decreasing.

Answer 1

The Correct Option is C

To determine the nature of the function \( f(x) = x^3 - 3x^2 + 4x \), we need to find the first derivative of the function and analyze its sign. First, differentiate \( f(x) \) with respect to \( x \): \[ f'(x) = \frac{d}{dx} \left( x^3 - 3x^2 + 4x \right) \] Using the power rule: \[ f'(x) = 3x^2 - 6x + 4 \] Now, we examine the sign of \( f'(x) \). Since the first derivative \( f'(x) = 3x^2 - 6x + 4 \) is a quadratic expression, we can find its discriminant to check whether the function is increasing or decreasing. The discriminant \( \Delta \) of the quadratic equation \( 3x^2 - 6x + 4 = 0 \) is: \[ \Delta = (-6)^2 - 4(3)(4) = 36 - 48 = -12 \] Since the discriminant is negative, the quadratic equation has no real roots, meaning \( f'(x) \) does not change sign. Now, we check the sign of \( f'(x) \) for one value of \( x \). For example, at \( x = 0 \): \[ f'(0) = 3(0)^2 - 6(0) + 4 = 4 \] Since \( f'(x)>0 \) for all \( x \), the function \( f(x) \) is strictly increasing on the real number line. Thus, the correct answer is \( \text{C} \).