Consider the following for the next items that follow: A function is defined by\(f(x)=\left|\begin{array}{ccc} x+1 & 2 & 3 \\ 2 & x+4 & 6 \\ 3 & 6 & x+9 \end{array}\right|\) .
The function is decreasing on :
Understanding the Determinant Function
The problem asks us to find the interval where the function \(f(x)\) is decreasing. The function is defined by a determinant:
\[f(x)=\left|\begin{array}{ccc} x+1 & 2 & 3 \\ 2 & x+4 & 6 \\ 3 & 6 & x+9 \end{array}\right|\]To determine where a function is decreasing, we need to find its first derivative, \(f'(x)\), and identify the intervals where \(f'(x) < 0\).
Evaluating the Determinant to Find f(x)
We evaluate the determinant to get an explicit expression for \(f(x)\). We can expand along the first row:
\[f(x) = (x+1)\left|\begin{array}{cc} x+4 & 6 \\ 6 & x+9 \end{array}\right| - 2\left|\begin{array}{cc} 2 & 6 \\ 3 & x+9 \end{array}\right| + 3\left|\begin{array}{cc} 2 & x+4 \\ 3 & 6 \end{array}\right|\]Now, we evaluate the 2x2 determinants:
- The first minor: \((x+4)(x+9) - 6 \times 6 = (x^2 + 9x + 4x + 36) - 36 = x^2 + 13x\)
- The second minor: \(2(x+9) - 6 \times 3 = 2x + 18 - 18 = 2x\)
- The third minor: \(2 \times 6 - 3(x+4) = 12 - 3x - 12 = -3x\)
Substitute these back into the expression for \(f(x)\):
\[f(x) = (x+1)(x^2 + 13x) - 2(2x) + 3(-3x)\] \[f(x) = (x^3 + 13x^2 + x^2 + 13x) - 4x - 9x\] \[f(x) = x^3 + 14x^2 + 13x - 13x\] \[f(x) = x^3 + 14x^2\]So, the function is \(f(x) = x^3 + 14x^2\).
Finding the Derivative f'(x)
Now we find the derivative of \(f(x)\) with respect to \(x\):
\[f'(x) = \frac{d}{dx}(x^3 + 14x^2)\] \[f'(x) = 3x^2 + 28x\]Determining Intervals of Decrease
A function is decreasing on an interval where its derivative \(f'(x)\) is negative. We need to find the values of \(x\) for which \(3x^2 + 28x < 0\).
First, find the critical points by setting \(f'(x) = 0\):
\[3x^2 + 28x = 0\]Factor out \(x\):
\[x(3x + 28) = 0\]This gives us two critical points:
- \(x = 0\)
- \(3x + 28 = 0 \Rightarrow 3x = -28 \Rightarrow x = -\frac{28}{3}\)
These critical points, \(x = -\frac{28}{3}\) and \(x = 0\), divide the number line into three intervals: \((-\infty, -\frac{28}{3})\), \((-\frac{28}{3}, 0)\), and \((0, \infty)\). We will test the sign of \(f'(x) = x(3x + 28)\) in each interval.
| Interval | Test Value \(x\) | \(x\) | \(3x + 28\) | \(f'(x) = x(3x + 28)\) | Sign of \(f'(x)\) | Behavior of \(f(x)\) |
|---|---|---|---|---|---|---|
| \((-\infty, -\frac{28}{3})\) | \(x = -10\) | Negative | \(3(-10) + 28 = -30 + 28 = -2\) (Negative) | \((\text{Negative}) \times (\text{Negative})\) | Positive (> 0) | Increasing |
| \((-\frac{28}{3}, 0)\) | \(x = -1\) | Negative | \(3(-1) + 28 = -3 + 28 = 25\) (Positive) | \((\text{Negative}) \times (\text{Positive})\) | Negative (< 0) | Decreasing |
| \((0, \infty)\) | \(x = 1\) | Positive | \(3(1) + 28 = 3 + 28 = 31\) (Positive) | \((\text{Positive}) \times (\text{Positive})\) | Positive (> 0) | Increasing |
The sign analysis shows that \(f'(x) < 0\) in the interval \((-\frac{28}{3}, 0)\). Since the function is continuous and differentiable, it is decreasing on the closed interval that includes the critical points where \(f'(x) = 0\). Thus, the function is decreasing on the interval \([-\frac{28}{3}, 0]\).
Conclusion on Decreasing Interval
Based on the analysis of the derivative \(f'(x)\), the function \(f(x)\) is decreasing on the interval where \(f'(x) \le 0\). This corresponds to the closed interval \([-\frac{28}{3}, 0]\).
Revision Table: Key Steps
| Step | Description | Result |
|---|---|---|
| 1 | Define the function \(f(x)\) using the determinant. | Determinant form provided. |
| 2 | Evaluate the determinant to find \(f(x)\) in terms of \(x\). | \(f(x) = x^3 + 14x^2\) |
| 3 | Calculate the first derivative \(f'(x)\). | \(f'(x) = 3x^2 + 28x\) |
| 4 | Find critical points by setting \(f'(x) = 0\). | \(x = -\frac{28}{3}, x = 0\) |
| 5 | Test the sign of \(f'(x)\) in intervals defined by critical points. | \(f'(x) < 0\) on \((-\frac{28}{3}, 0)\) |
| 6 | Identify the interval(s) where \(f'(x) \le 0\) for decreasing behavior. | \([-\frac{28}{3}, 0]\) |
Additional Information: Increasing and Decreasing Functions
For a function \(f(x)\) that is continuous on \([a, b]\) and differentiable on \((a, b)\):
- If \(f'(x) > 0\) for all \(x\) in \((a, b)\), then \(f(x)\) is increasing on \([a, b]\).
- If \(f'(x) < 0\) for all \(x\) in \((a, b)\), then \(f(x)\) is decreasing on \([a, b]\).
- If \(f'(x) = 0\) for all \(x\) in \((a, b)\), then \(f(x)\) is constant on \([a, b]\).
Critical points are points where \(f'(x) = 0\) or \(f'(x)\) is undefined. These points are important because they often mark the transition between intervals where the function is increasing or decreasing.
The determinant of a matrix can define a function, and calculus concepts like derivatives can be applied to analyze its behavior, such as finding intervals of increase or decrease, local extrema, concavity, etc.
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