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If the function f(x) = 2x3 - 9ax2 + 12a2x +1 has a local maximum at x = x1 and a local minimum at x = x2 such that x2 = x21, then a is equal t o _______ .
Correct answer is '2'. Can you explain this answer?
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If the function f(x) = 2x3 - 9ax2 + 12a2x +1 has a local maximum at x ...
Given f(x) = 2x3 - 9ax2 + 12a2x +1
for max or min, f (x) = 0
⇒ 6x2-18ax + 12a2 = 0
for max or min, f (x) = 0
⇒ 6x2-18ax + 12a2 = 0
Most Upvoted Answer
If the function f(x) = 2x3 - 9ax2 + 12a2x +1 has a local maximum at x ...
Given Information:
- Function: f(x) = 2x^3 - 9ax^2 + 12a^2x + 1
- Local maximum at x = x1
- Local minimum at x = x2
- x2 = x1^2
Explanation:
To find the value of 'a' that satisfies the given conditions, we need to analyze the behavior of the function f(x).
Finding the Derivative:
First, let's find the derivative of the function f(x) with respect to x. This will help us determine the critical points and the behavior of the function.
f'(x) = d/dx (2x^3 - 9ax^2 + 12a^2x + 1)
= 6x^2 - 18ax + 12a^2
Finding the Critical Points:
To find the critical points, we need to set the derivative equal to zero and solve for x.
6x^2 - 18ax + 12a^2 = 0
Case 1: x1 is a critical point (local maximum):
If x1 is a local maximum, then the derivative at x1 is equal to zero and the second derivative at x1 is negative.
f'(x1) = 0
6x1^2 - 18ax1 + 12a^2 = 0
f''(x1) < />
d^2/dx^2 (6x^2 - 18ax + 12a^2) < />
12x1 - 18a < />
Solving the above two equations simultaneously, we can find the value of x1 and 'a'. However, we are given that x2 = x1^2, so let's substitute x1^2 for x2 in the equations.
Case 2: x2 is a critical point (local minimum):
If x2 is a local minimum, then the derivative at x2 is equal to zero and the second derivative at x2 is positive.
f'(x2) = 0
6x2^2 - 18ax2 + 12a^2 = 0
f''(x2) > 0
d^2/dx^2 (6x^2 - 18ax + 12a^2) > 0
12x2 - 18a > 0
Since x2 = x1^2, let's substitute x1^2 for x2 in the equations.
Substituting x1^2 for x2:
- In the equation 6x1^2 - 18ax1 + 12a^2 = 0, replace x2 with x1^2:
6x1^2 - 18ax1 + 12a^2 = 0
- In the equation 12x2 - 18a > 0, replace x2 with x1^2:
12x1^2 - 18a > 0
Simplifying the Equations:
Now, let's simplify the equations to solve for 'a'.
Equation 1: 6x1^2 - 18ax1 +
- Function: f(x) = 2x^3 - 9ax^2 + 12a^2x + 1
- Local maximum at x = x1
- Local minimum at x = x2
- x2 = x1^2
Explanation:
To find the value of 'a' that satisfies the given conditions, we need to analyze the behavior of the function f(x).
Finding the Derivative:
First, let's find the derivative of the function f(x) with respect to x. This will help us determine the critical points and the behavior of the function.
f'(x) = d/dx (2x^3 - 9ax^2 + 12a^2x + 1)
= 6x^2 - 18ax + 12a^2
Finding the Critical Points:
To find the critical points, we need to set the derivative equal to zero and solve for x.
6x^2 - 18ax + 12a^2 = 0
Case 1: x1 is a critical point (local maximum):
If x1 is a local maximum, then the derivative at x1 is equal to zero and the second derivative at x1 is negative.
f'(x1) = 0
6x1^2 - 18ax1 + 12a^2 = 0
f''(x1) < />
d^2/dx^2 (6x^2 - 18ax + 12a^2) < />
12x1 - 18a < />
Solving the above two equations simultaneously, we can find the value of x1 and 'a'. However, we are given that x2 = x1^2, so let's substitute x1^2 for x2 in the equations.
Case 2: x2 is a critical point (local minimum):
If x2 is a local minimum, then the derivative at x2 is equal to zero and the second derivative at x2 is positive.
f'(x2) = 0
6x2^2 - 18ax2 + 12a^2 = 0
f''(x2) > 0
d^2/dx^2 (6x^2 - 18ax + 12a^2) > 0
12x2 - 18a > 0
Since x2 = x1^2, let's substitute x1^2 for x2 in the equations.
Substituting x1^2 for x2:
- In the equation 6x1^2 - 18ax1 + 12a^2 = 0, replace x2 with x1^2:
6x1^2 - 18ax1 + 12a^2 = 0
- In the equation 12x2 - 18a > 0, replace x2 with x1^2:
12x1^2 - 18a > 0
Simplifying the Equations:
Now, let's simplify the equations to solve for 'a'.
Equation 1: 6x1^2 - 18ax1 +
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If the function f(x) = 2x3 - 9ax2 + 12a2x +1 has a local maximum at x = x1 and a local minimum at x = x2 such that x2 = x21, then a is equal t o _______ .Correct answer is '2'. Can you explain this answer?
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