Problem 33 - Precalculus Mathematics for Calculus

Question

Find the maximum or minimum value of the function. $$f(x)=x^{2}+x+1$$

Step-by-Step Solution

Answer

The minimum value of the function is \(\frac{3}{4}\).

Step 1: Identify the Type of Quadratic Function

The function given is a quadratic function, written as \(f(x) = ax^2 + bx + c\). In this case, \(a = 1\), \(b = 1\), and \(c = 1\). Since \(a > 0\), the parabola opens upwards, indicating a minimum point exists.

Step 2: Determine the Vertex Formula

The vertex of a quadratic function \(f(x) = ax^2 + bx + c\) is found using the formula \(x = -\frac{b}{2a}\). This x-value gives the axis of symmetry for the parabola.

Step 3: Calculate the x-coordinate of the Vertex

Substitute the values of \(a\) and \(b\) into the vertex formula: \(x = -\frac{1}{2 \cdot 1} = -\frac{1}{2}\). This is the x-coordinate of the vertex where the minimum value occurs.

Step 4: Find the Minimum Value of the Function

Substitute \(x = -\frac{1}{2}\) into the function to find the minimum value: \[f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1\]\[= \frac{1}{4} - \frac{1}{2} + 1\]\[= \frac{1}{4} - \frac{2}{4} + \frac{4}{4}\]\[= \frac{3}{4}\].Thus, the minimum value of the function is \(\frac{3}{4}\).

Key Concepts

Vertex Formula

Quadratic functions often take the form of \( f(x) = ax^2 + bx + c \). A crucial part of understanding these functions is figuring out where they reach their extreme values, which is where the **vertex formula** comes into play. The vertex is either the highest or lowest point of a parabola, and it's found using the formula \( x = -\frac{b}{2a} \). This formula defines the x-coordinate of the vertex. For the given quadratic function \( f(x) = x^2 + x + 1 \), the coefficients are identified as \( a = 1 \) and \( b = 1 \). By substituting these values into the vertex formula, we calculate the x-value of the vertex as \( x = -\frac{1}{2} \). This single x-value points us to the axis of symmetry about which the parabola is mirrored. Understanding the vertex formula is fundamental for graphing parabolas and determining their extreme values.

Minimum Value

In quadratic functions like \( f(x) = x^2 + x + 1 \), identifying the minimum or maximum value is essential to understanding the parabolic shape. Since the coefficient \( a = 1 \) is positive, this tells us that the parabola opens upwards, indicative of a minimum point. The minimum value of a quadratic function occurs at the vertex. Here, we already identified that the x-coordinate of the vertex is \( x = -\frac{1}{2} \).

• Substitute this x-value back into the function to find the y-coordinate, which represents the minimum value:

\[ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 \]\[ = \frac{1}{4} - \frac{1}{2} + 1 \]\[ = \frac{3}{4} \]Thus, the minimum value of this quadratic function is \( \frac{3}{4} \). It's this minimum point on the graph that represents the lowest y-coordinate the function can reach.

Parabola Opening Upwards

Understanding how a parabola opens is crucial in predicting the behavior of the quadratic function. The direction in which a parabola opens is determined by the coefficient \( a \) in the quadratic function \( f(x) = ax^2 + bx + c \).

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), the parabola opens downwards.

For the function \( f(x) = x^2 + x + 1 \), since \( a = 1 \), the parabola opens upwards. This opening direction implies that the parabola has a minimum point. Knowing how the parabola opens aids in anticipating where the function reaches its extremum and what type of extreme value it holds—either a maximum or a minimum. For upward-opening parabolas like this one, you will look for the minimum value at the vertex, guiding your graphing efforts and offering insight into the range of the function.