Problem 29 - Algebra and Trigonometry

Question

\(29-38=\) 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 Function

The function given is \( f(x) = x^2 + x + 1 \). This is a quadratic function, which is a type of polynomial function of degree 2. Quadratic functions are expressed in the standard form \( ax^2 + bx + c \), where \( a = 1 \), \( b = 1 \), and \( c = 1 \) in this case.

Step 2: Determine the Direction of the Parabola

Because the coefficient of \( x^2 \) (\( a \)) is positive (\( a = 1 \)), the parabola opens upwards. This means that the function will achieve a minimum value at its vertex.

Step 3: Find the Vertex

For a quadratic function \( f(x) = ax^2 + bx + c \), the \( x \)-coordinate of the vertex is given by \( -\frac{b}{2a} \). Substituting \( a = 1 \) and \( b = 1 \) into the formula gives \( x = -\frac{1}{2 \times 1} = -\frac{1}{2} \).

Step 4: Calculate the Minimum Value

To find the minimum value of the function, substitute \( x = -\frac{1}{2} \) into the function. Thus, \( f(-\frac{1}{2}) = (-\frac{1}{2})^2 + (-\frac{1}{2}) + 1 \). Calculating, we have \( f(-\frac{1}{2}) = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4} \).

Step 5: Conclusion

The minimum value of the function \( f(x) = x^2 + x + 1 \) is \( \frac{3}{4} \), and this minimum value occurs at \( x = -\frac{1}{2} \).

Key Concepts

Vertex of a Parabola

When discussing quadratic functions, the vertex of a parabola is a key concept to understand. A parabola is the U-shaped curve that graphs a quadratic function. The vertex is the special point where the parabola changes direction. For an upward opening parabola, like in our case, it reaches its lowest point at the vertex. The location of this vertex can be determined using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \]For the quadratic function \( f(x) = x^2 + x + 1 \), the coefficients are \( a = 1 \), \( b = 1 \), and \( c = 1 \). Substituting these values into the vertex formula gives:\[ x = -\frac{1}{2 \times 1} = -\frac{1}{2}\]Thus, the parabola's vertex lies at \( x = -\frac{1}{2} \). This is a crucial feature as it helps in determining the parabola's minimum or maximum value.

Minimum Value of a Function

Every quadratic function represented by a parabola either has a highest or lowest point, which corresponds to the function’s maximum or minimum value. For parabolas that open upwards, like in our function \( f(x) = x^2 + x + 1 \), they have a minimum value at their vertex.The minimum value can be found by plugging the x-coordinate of the vertex back into the original function. We've found the x-coordinate of the vertex to be \( x = -\frac{1}{2} \). Now, let's find the corresponding \( f(x) \):\[ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 \]Calculating step by step gives:

• \((-\frac{1}{2})^2 = \frac{1}{4}\)
• Adding \(-\frac{1}{2} \) yields \( \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \)
• Finally, adding \(1\) gives \( -\frac{1}{4} + 1 = \frac{3}{4} \)

Thus, the minimum value of the function is \( \frac{3}{4} \). This clear point indicates where the parabolic curve reaches its lowest level on the graph, making it a pivotal aspect when analyzing the function.

Polynomial Function of Degree 2

Quadratic functions are a specific type of polynomial function. They are termed as polynomial functions of degree 2 since the highest power of the variable \( x \) is 2. In general, these functions can be expressed in the form:\[ f(x) = ax^2 + bx + c \]In our example, \( f(x) = x^2 + x + 1 \), where:

• The coefficient \( a = 1 \) determines the direction of the parabola (since it's positive, our parabola opens upwards).
• The coefficient \( b = 1 \) affects the tilt or direction along the x-axis.
• The constant term \( c = 1 \) moves the entire graph up or down in the y-axis.

Such simplicity in a second-degree polynomial function presents a straightforward pattern. This pattern is easy to visualize as the parabola graphically represents how quadratic combinations of coefficients affect shapes and directions. Understanding each element's role enables a clearer comprehension of the behavior and properties of these functions.