Problem 55 - Calculus Early Transcendentals

Question

Let \(f(x)=a x^{2}+b x+c,\) where \(a>0 .\) Prove that \(f(x) \geq 0\) for all \(x\) if and only if \(b^{2}-4 a c \leq 0 .\) [Hint: Find the minimum of \(f(x) .]\)

Step-by-Step Solution

Answer

The function \( f(x) \geq 0 \) for all \( x \) if and only if \( b^2 - 4ac \leq 0 \).

Step 1: Understanding the Quadratic Function

A quadratic function of the form \( f(x) = ax^2 + bx + c \) is a parabola. Since \( a > 0 \), it opens upwards, which means it has a minimum point.

Step 2: Find the Minimum of the Function

To find the minimum point of the quadratic function \( f(x) = ax^2 + bx + c \), we can use the vertex formula for the x-coordinate, \( x = -\frac{b}{2a} \).

Step 3: Calculate the Minimum Value

Substitute \( x = -\frac{b}{2a} \) into \( f(x) \):\[f\left(-\frac{b}{2a}\right) = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c.\]Simplifying gives:\[f\left(-\frac{b}{2a}\right) = \frac{b^2}{4a} - \frac{b^2}{2a} + c = \frac{b^2}{4a} - \frac{2b^2}{4a} + c = \frac{b^2 - 4ac}{4a}.\]

Step 4: Determine the Condition for Non-Negativity

For \( f(x) \geq 0 \) for all \( x \), the minimum value \( \frac{b^2 - 4ac}{4a} \) must be non-negative. Therefore, \( \frac{b^2 - 4ac}{4a} \geq 0 \) simplifies to \( b^2 - 4ac \leq 0 \).

Step 5: Conclusion

Hence, the original statement is proven: The function \( f(x) = ax^2 + bx + c \) satisfies \( f(x) \geq 0 \) for all \( x \) if and only if \( b^2 - 4ac \leq 0 \).

Key Concepts

Parabola

A quadratic function appears as a parabola on a graph. This form of equation, written as \( f(x) = ax^2 + bx + c \), has a distinct U-shape. When the coefficient \( a \) is greater than zero, the parabola opens upwards. This upward opening is crucial because it signifies that there is a minimum point somewhere along the curve. The lowest point of this upward-opening parabola is called the vertex. Understanding the nature of a parabola is key. Here are some important features to remember about parabolas:

• If \( a > 0 \), the parabola opens upwards and the vertex represents a minimum point.
• The axis of symmetry passes through the vertex and divides the parabola into two mirror images.
• The shape and width of the parabola depend on the value of \( a \). A larger \( a \) makes the parabola narrower, while a smaller \( a \) makes it wider.

Knowing these properties will help you visualize and solve problems involving quadratic functions with confidence.

Vertex Formula

To pinpoint the minimum or maximum position of a parabola, you need to find its vertex. For a quadratic function \( f(x) = ax^2 + bx + c \), the vertex formula is essential. The x-coordinate of the vertex is calculated with the formula \( x = -\frac{b}{2a} \). Once this x-value is found, it can be substituted back into the original equation to find the corresponding y-coordinate, which gives the minimum or maximum value of the function. Here's a step-by-step:

• Identify the coefficients \( a \), \( b \), and \( c \) in the quadratic expression.
• Apply the vertex formula \( x = -\frac{b}{2a} \) to find the x-coordinate of the vertex.
• Substitute \( x = -\frac{b}{2a} \) into the function \( f(x) \) to determine the minimum or maximum value (the y-coordinate).
  • This means finding \( f\left(-\frac{b}{2a}\right) \).

Understanding how to use the vertex formula is crucial to solving many problems involving quadratic equations. It provides a direct method to locate the peak or valley of the parabola.

Discriminant

The discriminant is an integral part of analyzing quadratic equations. It is calculated using the formula \( b^2 - 4ac \), found inside the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). The value of the discriminant tells us about the nature of the roots of a quadratic equation and helps in determining the characteristics of the parabola:

• If the discriminant is positive \((b^2 - 4ac > 0)\), the quadratic equation has two distinct real roots and the parabola intersects the x-axis at two points.
• If the discriminant is zero \((b^2 - 4ac = 0)\), the parabola touches the x-axis at one point. This means there is exactly one real root, and the vertex sits on the x-axis.
• If the discriminant is negative \((b^2 - 4ac < 0)\), the equation has no real roots and the parabola does not intersect the x-axis. In such cases, the graph of the function stays above or below the x-axis, depending on the direction in which the parabola opens.

For the specific problem we're considering, the statement \( b^2 - 4ac \leq 0 \) indicates that the quadratic function \( f(x) = ax^2 + bx + c \) is always non-negative, or remains above the x-axis. This is essential for understanding which solutions pertain to continuous non-negativity.