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 attains local minimum value at :
x = 0
Understanding the Function and the Problem
The question asks us to find the value of \(x\) at which the given function \(f(x)\) attains a local minimum value. The function is defined as a 3x3 determinant:
\[f(x)=\left|\begin{array}{ccc} x+1 & 2 & 3 \\ 2 & x+4 & 6 \\ 3 & 6 & x+9 \end{array}\right|\]
To find the local minimum of this function, we first need to express \(f(x)\) as a polynomial in \(x\). Then, we will use the principles of calculus, specifically derivatives, to find the critical points and classify them as local maxima, minima, or saddle points.
Step-by-Step Calculation of the Determinant \(f(x)\)
We can expand the determinant 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 2x2 determinant is \((x+4)(x+9) - (6)(6) = (x^2 + 9x + 4x + 36) - 36 = x^2 + 13x\).
- The second 2x2 determinant is \((2)(x+9) - (6)(3) = (2x + 18) - 18 = 2x\).
- The third 2x2 determinant is \((2)(6) - (x+4)(3) = 12 - (3x + 12) = 12 - 3x - 12 = -3x\).
Substitute these values back into the expression for \(f(x)\):
\[f(x) = (x+1)(x^2 + 13x) - 2(2x) + 3(-3x)\]
Expand and simplify the expression:
\[f(x) = (x^3 + 13x^2 + x^2 + 13x) - 4x - 9x\]
\[f(x) = x^3 + 14x^2 + 13x - 13x\]
\[f(x) = x^3 + 14x^2\]
So, the function is \(f(x) = x^3 + 14x^2\).
Finding Critical Points for Local Extrema
To find the critical points where local extrema might occur, we need to compute the first derivative of \(f(x)\) with respect to \(x\) and set it equal to zero.
\[f'(x) = \frac{d}{dx}(x^3 + 14x^2)\]
\[f'(x) = 3x^2 + 28x\]
Set \(f'(x) = 0\):
\[3x^2 + 28x = 0\]
Factor out \(x\):
\[x(3x + 28) = 0\]
This gives us two critical points:
- \(x = 0\)
- \(3x + 28 = 0 \implies 3x = -28 \implies x = -\frac{28}{3}\)
Classifying Critical Points using the Second Derivative Test
To determine whether these critical points correspond to a local minimum or maximum, we use the second derivative test. First, we find the second derivative of \(f(x)\).
\[f''(x) = \frac{d}{dx}(f'(x)) = \frac{d}{dx}(3x^2 + 28x)\]
\[f''(x) = 6x + 28\]
Now, we evaluate \(f''(x)\) at each critical point:
- At \(x = 0\):
\[f''(0) = 6(0) + 28 = 28\] Since \(f''(0) = 28 > 0\), the function has a local minimum at \(x = 0\). - At \(x = -\frac{28}{3}\):
\[f''\left(-\frac{28}{3}\right) = 6\left(-\frac{28}{3}\right) + 28 = 2(-28) + 28 = -56 + 28 = -28\] Since \(f''\left(-\frac{28}{3}\right) = -28 < 0\), the function has a local maximum at \(x = -\frac{28}{3}\).
Conclusion
Based on the second derivative test, the function \(f(x) = x^3 + 14x^2\) attains a local minimum value at \(x = 0\).
Comparing this result with the given options, we find that \(x=0\) is indeed one of the options.
| Critical Point \(x\) | Value of \(f''(x)\) | Classification |
|---|---|---|
| \(0\) | \(28\) | Local Minimum |
| \(-\frac{28}{3}\) | \(-28\) | Local Maximum |
Revision Table: Finding Local Extrema
| Step | Description | Applied to \(f(x) = x^3 + 14x^2\) |
|---|---|---|
| 1 | Find the first derivative, \(f'(x)\). | \(f'(x) = 3x^2 + 28x\) |
| 2 | Set \(f'(x) = 0\) and solve for \(x\) to find critical points. | \(3x^2 + 28x = 0 \implies x(3x+28)=0 \implies x=0, x=-\frac{28}{3}\) |
| 3 | Find the second derivative, \(f''(x)\). | \(f''(x) = 6x + 28\) |
| 4 | Evaluate \(f''(x)\) at each critical point. | \(f''(0) = 28\), \(f''(-\frac{28}{3}) = -28\) |
| 5 | Classify points: If \(f''(x) > 0\), local minimum; If \(f''(x) < 0\), local maximum; If \(f''(x) = 0\), test is inconclusive. | At \(x=0\), \(f''(0) > 0 \implies\) Local Minimum At \(x=-\frac{28}{3}\), \(f''(-\frac{28}{3}) < 0 \implies\) Local Maximum |
Additional Information: Determinants and Calculus
This problem combines concepts from linear algebra (determinants) and differential calculus. Understanding how to calculate determinants is the first crucial step. The determinant of a 3x3 matrix \(\left|\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}\right|\) can be calculated as \(a(ei-fh) - b(di-fg) + c(dh-eg)\).
Once the function is expressed in a standard polynomial form, finding local extrema is a typical application of differential calculus. The critical points are candidates for local extrema. The second derivative test provides a straightforward way to classify these points by examining the concavity of the function at those points. A positive second derivative indicates concave up, which corresponds to a local minimum. A negative second derivative indicates concave down, corresponding to a local maximum.
In some cases where the second derivative test is inconclusive (\(f''(x) = 0\)), or if it's easier, the first derivative test can be used. The first derivative test involves examining the sign of \(f'(x)\) on either side of a critical point to see if the function is changing from increasing to decreasing (local maximum) or decreasing to increasing (local minimum).
Important Questions from Applications of Derivatives
The function is decreasing on :
What is the maximum value of y?
What is the maximum value of xy ?
Consider the following statements:
1. \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) is an increasing function on [0, ∞).
2. \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) is an increasing function on (-∞, ∞).
Which of the above statements is/are correct?
\({\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}\) where x ϵ R
At what value of x does f(x) attain minimum value?
