Mathematics Exam > Mathematics Questions > Consider the f(x, y) = x2+ y2– a. For w...
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Consider the f(x, y) = x2 + y2 – a. For what values of a do we have critical points for the function
- a)independent of a
- b)for any real number except zero
- c)a ∊ (0, +∞)
- d)a ∊ (-1, 1)
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Consider the f(x, y) = x2+ y2– a. For what values of a do we hav...
Consider
fx = 2x
and
fy = 2y
There is no a here. Thus, independent of a.
Most Upvoted Answer
Consider the f(x, y) = x2+ y2– a. For what values of a do we hav...
Understanding the Function
The function in question is f(x, y) = x^2 + y^2 - a. To determine the critical points, we first look for where the gradient (∇f) is equal to zero.
Finding the Gradient
To find critical points, we compute the partial derivatives:
- ∂f/∂x = 2x
- ∂f/∂y = 2y
Setting these derivatives to zero gives us:
- 2x = 0 → x = 0
- 2y = 0 → y = 0
Thus, the only critical point is at (0, 0).
Analyzing the Function Value
At the critical point (0, 0):
- f(0, 0) = 0 + 0 - a = -a
For critical points to exist, a must be a real number that defines the value of the function at the critical point. The critical point (0, 0) will exist regardless of the value of 'a'.
Conclusion on Values of 'a'
- The critical point is independent of 'a'. Therefore, the function will have critical points for any real number 'a'.
Correct Answer
The correct answer is option A: independent of a. This means that no restrictions are placed on 'a' for the existence of critical points.
Critical points exist in the context of the function's behavior, not tied to a specific value of 'a', thus highlighting the independence of the parameter in this scenario.
The function in question is f(x, y) = x^2 + y^2 - a. To determine the critical points, we first look for where the gradient (∇f) is equal to zero.
Finding the Gradient
To find critical points, we compute the partial derivatives:
- ∂f/∂x = 2x
- ∂f/∂y = 2y
Setting these derivatives to zero gives us:
- 2x = 0 → x = 0
- 2y = 0 → y = 0
Thus, the only critical point is at (0, 0).
Analyzing the Function Value
At the critical point (0, 0):
- f(0, 0) = 0 + 0 - a = -a
For critical points to exist, a must be a real number that defines the value of the function at the critical point. The critical point (0, 0) will exist regardless of the value of 'a'.
Conclusion on Values of 'a'
- The critical point is independent of 'a'. Therefore, the function will have critical points for any real number 'a'.
Correct Answer
The correct answer is option A: independent of a. This means that no restrictions are placed on 'a' for the existence of critical points.
Critical points exist in the context of the function's behavior, not tied to a specific value of 'a', thus highlighting the independence of the parameter in this scenario.
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