While minimizing the function f(x), necessary and sufficient conditions for a point, x0 to be a minima are:
When we want to find the lowest points of a function, also known as its local minima, we use a set of conditions involving the derivatives of the function. These conditions help us identify specific points on the function's graph where a minimum value occurs. Understanding these necessary and sufficient conditions is crucial for optimization problems in calculus.
The first step in finding a local minimum for a function \(f(x)\) at a point \(x_0\) is to identify its critical points. A critical point is where the function's slope is zero, meaning the tangent line to the curve at that point is horizontal. This is expressed by the first derivative being equal to zero:
$$f'\left( {{x_0}} \right) = 0$$
This condition is necessary because if the first derivative is not zero, the function is either increasing or decreasing at that point, and thus it cannot be a local extremum (minimum or maximum). However, this condition alone is not sufficient, as a critical point could also be a local maximum or an inflection point (saddle point).
Once we have identified a critical point \(x_0\) where \(f'\left( {{x_0}} \right) = 0\), we need an additional condition to confirm if it's indeed a local minimum. This is where the second derivative comes into play. The second derivative tells us about the concavity of the function at that point.
For a point \(x_0\) to be a local minimum, the function must be "concave up" at \(x_0\). This means the curve holds water, visually speaking. Mathematically, this is indicated by a positive second derivative:
$$f''\left( {{x_0}} \right) > 0$$
If the second derivative at \(x_0\) is positive, it confirms that the critical point is a local minimum. If \(f''\left( {{x_0}} \right) < 0\), it's a local maximum. If \(f''\left( {{x_0}} \right) = 0\), the second derivative test is inconclusive, and higher-order derivatives or other methods might be needed.
Combining both the necessary and sufficient conditions, for a point \(x_0\) to be a local minimum of the function \(f(x)\), two conditions must simultaneously hold true:
Let's evaluate each given option based on these principles for minimizing a function:
| Option | First Derivative Condition | Second Derivative Condition | Conclusion for Minima |
|---|---|---|---|
| 1 | \(f'\left( {{x_0}} \right) > 0\) | \(f''\left( {{x_0}} \right) = 0\) | Incorrect. \(f'\left( {{x_0}} \right) > 0\) means the function is increasing, not at a critical point. Thus, it cannot be a minimum. |
| 2 | \(f'\left( {{x_0}} \right) < 0\) | \(f''\left( {{x_0}} \right) = 0\) | Incorrect. \(f'\left( {{x_0}} \right) < 0\) means the function is decreasing, not at a critical point. Thus, it cannot be a minimum. |
| 3 | \(f'\left( {{x_0}} \right) = 0\) | \(f''\left( {{x_0}} \right) = 0\) | Incorrect. While \(f'\left( {{x_0}} \right) = 0\) identifies a critical point, \(f''\left( {{x_0}} \right) = 0\) means the second derivative test is inconclusive. The point could be a minimum, maximum, or an inflection point. |
| 4 | \(f'\left( {{x_0}} \right) = 0\) | \(f''\left( {{x_0}} \right) > 0\) | Correct. This option satisfies both the necessary and sufficient conditions for a local minimum. The first derivative is zero (critical point), and the second derivative is positive (concave up), confirming it is a minimum. |
Therefore, for a point \(x_0\) to be a local minimum of the function \(f(x)\), the necessary and sufficient conditions are:
For the equation \(\frac{{dy}}{{dx}} + 7{x^2}y = 0\) , if y(0) = \(\frac{{3}}{{7}}\) , then the value of y(1) is
The differential equation \(\frac{{dy}}{{dx}} + 4y = 5\) is valid in the domain 0 ≤ x ≤ 1 with y (0) = 2.25 The solution of the differential equation is
The derivative of f(x) = cos(x) can be estimated using the approximation \(f'\left( x \right) = \frac{{f\left( {x + h} \right) - f\left( {x - h} \right)}}{{2h}}\) . The percentage error is calculated as \(\left( {\frac{{Exact\;value - Approximate\;value}}{{Exact\;value}}} \right) \times 100\). The percentage error in the derivative of f(x) at x = π/6 radian, choosing h = 0.1 radian, is
The general solution of the differential equation \(\frac{{dy}}{{dx}} = \cos \left( {x + y} \right)\), with c as a constant, is
Which one of the following is the general solution of the first order differential equation
\(\frac{{dy}}{{dx}} = {\left( {x + y - 1} \right)^2}\) , where x, y are real?