Optimum Value Determination for Quadratic Functions

The optimum value of a function refers to its minimum or maximum value. For a quadratic function in the standard form \(f(x) = ax^2 + bx + c\), the graph is a parabola. The nature of the optimum value (whether it's a minimum or a maximum) depends on the coefficient of the \(x^2\) term, 'a'.

• If \(a > 0\), the parabola opens upwards, and the function has a minimum value.
• If \(a < 0\), the parabola opens downwards, and the function has a maximum value.

Understanding the Given Quadratic Function

The function given is \(f(x) = x^2 - 4x + 2\).

Comparing this to the standard form \(f(x) = ax^2 + bx + c\), we can identify the coefficients:

• \(a = 1\)
• \(b = -4\)
• \(c = 2\)

Since the value of \(a = 1\), which is greater than 0 (\(a > 0\)), the parabola opens upwards. This means the function will have a minimum value.

Method 1: Using the Vertex Formula to Find Optimum Value

The x-coordinate of the vertex of a parabola, which represents the point where the function reaches its minimum or maximum value, can be found using the formula:

$$x = -\frac{b}{2a}$$

Substitute the values of \(a = 1\) and \(b = -4\) into the formula:

$$x = -\frac{(-4)}{2(1)} = \frac{4}{2} = 2$$

Now, to find the optimum (minimum) value of the function, substitute this x-coordinate back into the original function \(f(x)\):

$$f(2) = (2)^2 - 4(2) + 2$$ $$f(2) = 4 - 8 + 2$$ $$f(2) = -4 + 2$$ $$f(2) = -2$$

Therefore, the minimum value of the function is \(-2\).

Method 2: Using Calculus (First Derivative Test)

For a function to reach an optimum value (minimum or maximum), its first derivative must be equal to zero at that point. This point is called a critical point.

First, find the first derivative of \(f(x) = x^2 - 4x + 2\):

$$f'(x) = \frac{d}{dx}(x^2 - 4x + 2)$$ $$f'(x) = 2x - 4$$

Next, set the first derivative to zero to find the critical x-value:

$$2x - 4 = 0$$ $$2x = 4$$ $$x = 2$$

Now, substitute this x-value back into the original function \(f(x)\) to find the optimum value:

$$f(2) = (2)^2 - 4(2) + 2$$ $$f(2) = 4 - 8 + 2$$ $$f(2) = -2$$

To confirm whether this is a minimum or maximum, we can use the second derivative test. Find the second derivative of \(f(x)\):

$$f''(x) = \frac{d}{dx}(2x - 4)$$ $$f''(x) = 2$$

Since \(f''(x) = 2\), which is a positive value (\(f''(x) > 0\)), it confirms that the critical point corresponds to a minimum value.

Conclusion on Optimum Value

Both methods consistently show that the optimum value of the function \(f(x) = x^2 - 4x + 2\) is \(-2\). As determined by the coefficient of the \(x^2\) term (\(a=1 > 0\)), this optimum value is a minimum.

Therefore, the optimum value is \(-2\) (minimum).