What is the maximum value of the expression \(\frac{1}{{{x^2}\; + \;5x\; + \;10}}?\)
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Finding the Maximum Value of a Rational Expression
To find the maximum value of the expression \(\frac{1}{{{x^2}\; + \;5x\; + \;10}}\), we need to analyze its structure. The expression is a fraction with a constant numerator (1) and a variable denominator (\({x^2}\; + \;5x\; + \;10\)).
For a fraction with a positive numerator, the maximum value is achieved when the denominator has its minimum positive value. If the denominator can be zero or negative, the fraction could be undefined or negative, but let's first find the minimum value of the denominator.
Minimizing the Denominator: A Quadratic Expression
The denominator is the quadratic expression \(D(x) = {x^2}\; + \;5x\; + \;10\). This is in the standard form \(ax^2 + bx + c\), where \(a=1\), \(b=5\), and \(c=10\).
Since the coefficient of the \(x^2\) term, \(a=1\), is positive (\(a > 0\)), the parabola represented by this quadratic opens upwards. An upward-opening parabola has a minimum value at its vertex.
Calculating the Minimum Value of the Denominator
The x-coordinate of the vertex of a parabola \(ax^2 + bx + c\) is given by the formula \(x = -\frac{b}{2a}\).
Substituting the values from our denominator \(D(x) = {x^2}\; + \;5x\; + \;10\):
\[x = -\frac{5}{2 \times 1} = -\frac{5}{2}\]
This means the minimum value of the denominator occurs when \(x = -\frac{5}{2}\).
Now, we substitute this value of \(x\) back into the denominator expression to find the minimum value:
\[\text{Minimum value of } D(x) = \left(-\frac{5}{2}\right)^2 + 5\left(-\frac{5}{2}\right) + 10\]
Let's calculate this step-by-step:
- Square \(-\frac{5}{2}\): \(\left(-\frac{5}{2}\right)^2 = \frac{(-5)^2}{2^2} = \frac{25}{4}\)
- Multiply 5 by \(-\frac{5}{2}\): \(5\left(-\frac{5}{2}\right) = -\frac{25}{2}\)
- The constant term is 10.
So, the minimum value is:
\[\text{Minimum value of } D(x) = \frac{25}{4} - \frac{25}{2} + 10\]
To combine these terms, we find a common denominator, which is 4:
\[\text{Minimum value of } D(x) = \frac{25}{4} - \frac{25 \times 2}{2 \times 2} + \frac{10 \times 4}{1 \times 4}\]
\[\text{Minimum value of } D(x) = \frac{25}{4} - \frac{50}{4} + \frac{40}{4}\]
\[\text{Minimum value of } D(x) = \frac{25 - 50 + 40}{4}\]
\[\text{Minimum value of } D(x) = \frac{-25 + 40}{4}\]
\[\text{Minimum value of } D(x) = \frac{15}{4}\]
The minimum value of the denominator \(x^2 + 5x + 10\) is \(\frac{15}{4}\).
Since the minimum value \(\frac{15}{4}\) is positive, the denominator is always positive for all real values of \(x\).
Finding the Maximum Value of the Expression
The original expression is \(\frac{1}{{{x^2}\; + \;5x\; + \;10}}\). Its maximum value occurs when the denominator is at its minimum positive value.
Maximum value of expression = \(\frac{1}{\text{Minimum value of denominator}}\)
Maximum value of expression = \(\frac{1}{\frac{15}{4}}\)
To divide by a fraction, we multiply by its reciprocal:
Maximum value of expression = \(1 \times \frac{4}{15} = \frac{4}{15}\)
Thus, the maximum value of the expression \(\frac{1}{{{x^2}\; + \;5x\; + \;10}}\) is \(\frac{4}{15}\).
| Step | Description | Calculation/Analysis |
|---|---|---|
| 1 | Identify the expression structure | \(\frac{1}{\text{Quadratic Denominator}}\) |
| 2 | Recognize condition for maximum | Maximize fraction \(\frac{1}{A}\) by minimizing denominator \(A\) (if \(A > 0\)) |
| 3 | Analyze the denominator | \(D(x) = x^2 + 5x + 10\) (Quadratic with \(a=1, b=5, c=10\)) |
| 4 | Find x-value for minimum of denominator | Vertex x-coordinate: \(x = -\frac{b}{2a} = -\frac{5}{2}\) |
| 5 | Calculate minimum value of denominator | Substitute \(x = -5/2\) into \(D(x)\): \(\left(-\frac{5}{2}\right)^2 + 5\left(-\frac{5}{2}\right) + 10 = \frac{25}{4} - \frac{25}{2} + 10 = \frac{15}{4}\) |
| 6 | Calculate maximum value of expression | \(\frac{1}{\text{Minimum Denominator}} = \frac{1}{15/4} = \frac{4}{15}\) |
Revision Table: Key Concepts Reviewed
| Concept | Description | Relevance to Problem |
|---|---|---|
| Quadratic Function | A function of the form \(f(x) = ax^2 + bx + c\). Its graph is a parabola. | The denominator \(x^2 + 5x + 10\) is a quadratic function. |
| Vertex of a Parabola | The lowest point (if \(a > 0\)) or highest point (if \(a < 0\)) on the parabola. | The minimum value of our quadratic denominator occurs at its vertex. |
| Vertex Formula (x-coord) | \(x = -\frac{b}{2a}\) gives the x-coordinate of the vertex. | Used to find the value of x where the denominator is minimized. |
| Maximizing a Fraction \(\frac{1}{A}\) | When the numerator is positive, the fraction is maximized when the denominator \(A\) is minimized (and positive). | Applied to find the maximum value of the given expression. |
Additional Information: Understanding Quadratic Minimums
For any quadratic function \(f(x) = ax^2 + bx + c\):
- If \(a > 0\), the parabola opens upwards, and the function has a minimum value. The minimum value is \(c - \frac{b^2}{4a}\), which occurs at \(x = -\frac{b}{2a}\).
- If \(a < 0\), the parabola opens downwards, and the function has a maximum value. The maximum value is \(c - \frac{b^2}{4a}\), which occurs at \(x = -\frac{b}{2a}\).
In our case, \(a=1\), \(b=5\), \(c=10\). The minimum value of the denominator can also be found using the formula \(c - \frac{b^2}{4a}\):
\[10 - \frac{5^2}{4 \times 1} = 10 - \frac{25}{4} = \frac{40}{4} - \frac{25}{4} = \frac{15}{4}\]
This confirms the minimum value of the denominator calculated earlier. Since the denominator's minimum value \(\frac{15}{4}\) is positive, the quadratic \(x^2 + 5x + 10\) is always positive for all real \(x\). Therefore, the reciprocal expression is well-defined for all real \(x\), and its maximum value is indeed \(1 / (\text{minimum denominator value})\).
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