Consider the function
\({\rm{f}}\left( {\rm{x}} \right) = \left| {\begin{array}{*{20}{c}} {{{\rm{x}}^3}}&{\sin {\rm{x}}}&{\cos {\rm{x}}}\\ 6&{ - 1}&0\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right|\) , where p is a constant What is the value of p for which f”(0) = 0?
Solving for p using Determinants and Derivatives
The problem asks for the value(s) of a constant \(p\) for which the second derivative of a function \(f(x)\) evaluated at \(x=0\) is equal to zero. The function \(f(x)\) is given as the determinant of a 3x3 matrix.
First, let's find the function \(f(x)\) by calculating the determinant of the given matrix:
\[ {\rm{f}}\left( {\rm{x}} \right) = \left| {\begin{array}{*{20}{c}} {{{\rm{x}}^3}}&{\sin {\rm{x}}}&{\cos {\rm{x}}}\\ 6&{ - 1}&0\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right| \]We can expand the determinant along the first row:
\[ f(x) = x^3 \left| {\begin{array}{*{20}{c}} { - 1}&0\\ {{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right| - \sin x \left| {\begin{array}{*{20}{c}} 6&0\\ {\rm{p}}&{{{\rm{p}}^3}} \end{array}} \right| + \cos x \left| {\begin{array}{*{20}{c}} 6&{ - 1}\\ {\rm{p}}&{{{\rm{p}}^2}} \end{array}} \right| \]Calculate the 2x2 determinants:
- \( \left| {\begin{array}{*{20}{c}} { - 1}&0\\ {{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right| = (-1)(p^3) - (0)(p^2) = -p^3 \)
- \( \left| {\begin{array}{*{20}{c}} 6&0\\ {\rm{p}}&{{{\rm{p}}^3}} \end{array}} \right| = (6)(p^3) - (0)(p) = 6p^3 \)
- \( \left| {\begin{array}{*{20}{c}} 6&{ - 1}\\ {\rm{p}}&{{{\rm{p}}^2}} \end{array}} \right| = (6)(p^2) - (-1)(p) = 6p^2 + p \)
Substitute these back into the expression for \(f(x)\):
\[ f(x) = x^3 (-p^3) - \sin x (6p^3) + \cos x (6p^2 + p) \] \[ f(x) = -p^3 x^3 - 6p^3 \sin x + (6p^2 + p) \cos x \]Finding the First Derivative \(f'(x)\)
Now, we need to find the first derivative of \(f(x)\) with respect to \(x\). Remember that \(p\) is a constant.
\[ f'(x) = \frac{d}{dx}(-p^3 x^3) + \frac{d}{dx}(-6p^3 \sin x) + \frac{d}{dx}((6p^2 + p) \cos x) \] \[ f'(x) = -p^3 (3x^2) - 6p^3 (\cos x) + (6p^2 + p) (-\sin x) \] \[ f'(x) = -3p^3 x^2 - 6p^3 \cos x - (6p^2 + p) \sin x \]Finding the Second Derivative \(f''(x)\)
Next, we find the second derivative by differentiating \(f'(x)\) with respect to \(x\).
\[ f''(x) = \frac{d}{dx}(-3p^3 x^2) + \frac{d}{dx}(-6p^3 \cos x) - \frac{d}{dx}((6p^2 + p) \sin x) \] \[ f''(x) = -3p^3 (2x) - 6p^3 (-\sin x) - (6p^2 + p) (\cos x) \] \[ f''(x) = -6p^3 x + 6p^3 \sin x - (6p^2 + p) \cos x \]Evaluating \(f''(0)\)
We need to evaluate \(f''(x)\) at \(x=0\). Substitute \(x=0\) into the expression for \(f''(x)\):
\[ f''(0) = -6p^3 (0) + 6p^3 \sin(0) - (6p^2 + p) \cos(0) \]We know that \( \sin(0) = 0 \) and \( \cos(0) = 1 \). Substitute these values:
\[ f''(0) = 0 + 6p^3 (0) - (6p^2 + p) (1) \] \[ f''(0) = 0 + 0 - (6p^2 + p) \] \[ f''(0) = -(6p^2 + p) \]Solving for p when \(f''(0) = 0\)
The problem states that \(f''(0) = 0\). So, we set the expression for \(f''(0)\) equal to zero and solve for \(p\):
\[ -(6p^2 + p) = 0 \] \[ 6p^2 + p = 0 \]This is a quadratic equation in \(p\). We can factor out \(p\) from the expression:
\[ p(6p + 1) = 0 \]For the product of two terms to be zero, at least one of the terms must be zero. This gives us two possible cases:
- Case 1: \(p = 0\)
- Case 2: \(6p + 1 = 0\)
Solving Case 2:
\[ 6p = -1 \] \[ p = -\frac{1}{6} \]Thus, the values of \(p\) for which \(f''(0) = 0\) are \(p = 0\) or \(p = -\frac{1}{6}\).
Summary of Results
The values of \(p\) that satisfy the condition \(f''(0) = 0\) are \(p = -\frac{1}{6}\) and \(p = 0\).
| Step | Description | Result |
|---|---|---|
| 1 | Calculate Determinant f(x) | \(f(x) = -p^3 x^3 - 6p^3 \sin x + (6p^2 + p) \cos x\) |
| 2 | Calculate First Derivative f'(x) | \(f'(x) = -3p^3 x^2 - 6p^3 \cos x - (6p^2 + p) \sin x\) |
| 3 | Calculate Second Derivative f''(x) | \(f''(x) = -6p^3 x + 6p^3 \sin x - (6p^2 + p) \cos x\) |
| 4 | Evaluate f''(0) | \(f''(0) = -(6p^2 + p)\) |
| 5 | Solve f''(0) = 0 for p | \(p = 0\) or \(p = -\frac{1}{6}\) |
Revision Table: Determinants and Derivatives
| Concept | Description | Key Formula/Rule |
|---|---|---|
| Determinant of 3x3 Matrix | A scalar value computed from the elements of a square matrix. It can be calculated by expanding along a row or column. | \( \left| {\begin{array}{*{20}{c}} a&b&c\\ d&e&f\\ g&h&i \end{array}} \right| = a(ei-fh) - b(di-fg) + c(dh-eg) \) |
| Derivative of \(x^n\) | The rate of change of \(x^n\) with respect to \(x\). | \( \frac{d}{dx}(x^n) = nx^{n-1} \) |
| Derivative of \(\sin x\) | The rate of change of \(\sin x\) with respect to \(x\). | \( \frac{d}{dx}(\sin x) = \cos x \) |
| Derivative of \(\cos x\) | The rate of change of \(\cos x\) with respect to \(x\). | \( \frac{d}{dx}(\cos x) = -\sin x \) |
| Constant Multiple Rule | The derivative of a constant times a function is the constant times the derivative of the function. | \( \frac{d}{dx}(cf(x)) = c \frac{d}{dx}(f(x)) \) |
| Sum/Difference Rule | The derivative of a sum or difference of functions is the sum or difference of their derivatives. | \( \frac{d}{dx}(f(x) \pm g(x)) = \frac{d}{dx}(f(x)) \pm \frac{d}{dx}(g(x)) \) |
| Second Derivative | The derivative of the first derivative, representing the rate of change of the rate of change of the function. | \( f''(x) = \frac{d}{dx}(f'(x)) \) |
Additional Information: Applications of Derivatives
Derivatives are fundamental tools in calculus with wide-ranging applications in science, engineering, economics, and many other fields. The first derivative \(f'(x)\) gives information about the slope of the tangent line to the graph of \(f(x)\) at a point, indicating whether the function is increasing, decreasing, or stationary.
The second derivative \(f''(x)\) provides information about the concavity of the graph of \(f(x)\). If \(f''(x) > 0\), the function is concave up; if \(f''(x) < 0\), the function is concave down. A point where the concavity changes is called an inflection point, and this often occurs where \(f''(x) = 0\).
Setting the second derivative to zero, as done in this problem (\(f''(0)=0\)), is often a step in finding inflection points or analyzing the behavior of the function around a specific point like \(x=0\). In this particular problem, the condition \(f''(0)=0\) leads to a specific constraint on the constant \(p\).
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\({\rm{f}}\left( {\rm{x}} \right) = \left| {\begin{array}{*{20}{c}} {{{\rm{x}}^3}}&{\sin {\rm{x}}}&{\cos {\rm{x}}}\\ 6&{ - 1}&0\\ {\rm{p}}&{{{\rm{p}}^2}}&{{{\rm{p}}^3}} \end{array}} \right|\) , where p is a constant
What is the value of f’(0)?
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