Question

Show that the given function \( f, f(x) = x^3 - 3x^2 + 4x, x \in \mathbb{R} \) is an increasing function in \( \mathbb{R} \).

Hint:

To quickly check if a quadratic function is always positive, check two conditions: 1. The coefficient of the \(x^2\) term (\(a\)) must be positive (\(a>0\)), which means the parabola opens upwards. 2. The discriminant (\(b^2 - 4ac\)) must be negative, which means the parabola has no real roots and never touches or crosses the x-axis. If both are true, the function is always positive.

Answer 1

Step 1: Understanding the Concept:
To determine if a function is increasing or decreasing over an interval, we use its first derivative. A function \( f(x) \) is strictly increasing on an interval if its first derivative, \( f'(x) \), is positive (\(f'(x)>0\)) for all \( x \) in that interval.
Step 2: Key Formula or Approach:
1. Find the first derivative of the function, \( f'(x) \).
2. Analyze the sign of \( f'(x) \) for all real numbers \( x \). We need to show that \( f'(x)>0 \) for all \( x \in \mathbb{R} \).
For a quadratic \(ax^2+bx+c\), if \(a>0\) and the discriminant \( \Delta = b^2 - 4ac<0 \), the quadratic is always positive.
Step 3: Detailed Explanation:
The given function is: \[ f(x) = x^3 - 3x^2 + 4x \] First, we find the derivative, \( f'(x) \): \[ f'(x) = \frac{d}{dx} (x^3 - 3x^2 + 4x) \] \[ f'(x) = 3x^2 - 6x + 4 \] Now, we need to show that this quadratic expression is always positive for all \( x \in \mathbb{R} \). We can do this by checking its discriminant (\( \Delta = b^2 - 4ac \)).
Here, \( a = 3, b = -6, c = 4 \). \[ \Delta = (-6)^2 - 4(3)(4) \] \[ \Delta = 36 - 48 \] \[ \Delta = -12 \] Since the discriminant \( \Delta = -12<0 \) and the leading coefficient \( a = 3>0 \), the quadratic \( 3x^2 - 6x + 4 \) is always positive for all real values of \( x \).
Alternatively, we can complete the square: \[ f'(x) = 3x^2 - 6x + 4 \] \[ f'(x) = 3(x^2 - 2x) + 4 \] \[ f'(x) = 3(x^2 - 2x + 1 - 1) + 4 \] \[ f'(x) = 3(x-1)^2 - 3 + 4 \] \[ f'(x) = 3(x-1)^2 + 1 \] Since \( (x-1)^2 \geq 0 \) for all \( x \in \mathbb{R} \), the minimum value of \( 3(x-1)^2 \) is 0.
Therefore, the minimum value of \( f'(x) \) is \( 0 + 1 = 1 \).
This means \( f'(x) \geq 1 \) for all \( x \in \mathbb{R} \).
Step 4: Final Answer:
Since \( f'(x)>0 \) for all \( x \in \mathbb{R} \), the function \( f(x) = x^3 - 3x^2 + 4x \) is an increasing function in \( \mathbb{R} \).