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The function f(x) = 2x3 – 3x2 – 36x + 2 has its maxima at  
  • a)
    x = – 2 only    
  • b)
    x = 0 only  
  • c)
    x = 3 only    
  • d)
    both x = –2 and x = 3 
Correct answer is option 'A'. Can you explain this answer?
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The function f(x) = 2x3 – 3x2 – 36x + 2 has its maxima at ...
Maxima of a Function

To determine the maxima of a function, we need to find the values of x where the derivative of the function is equal to zero or does not exist.

Given function: f(x) = 2x^3 - 3x^2 + 36x - 2

Finding the derivative of the function:

f'(x) = d/dx (2x^3 - 3x^2 + 36x - 2)
= 6x^2 - 6x + 36

Setting f'(x) equal to zero and solving for x:

6x^2 - 6x + 36 = 0

Dividing the equation by 6:

x^2 - x + 6 = 0

Using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

a = 1, b = -1, c = 6

x = (1 ± √((-1)^2 - 4(1)(6))) / (2(1))
x = (1 ± √(1 - 24)) / 2
x = (1 ± √(-23)) / 2

Since the expression inside the square root is negative, there are no real solutions for x. Therefore, f'(x) does not equal zero.

Since f'(x) does not equal zero, there are no critical points where the function reaches its maximum or minimum.

Therefore, the function does not have any maxima or minima.

Answer: None of the given options (A, B, C, D) is correct.
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The function f(x) = 2x3 – 3x2 – 36x + 2 has its maxima at a)x = – 2 only b)x = 0 only c)x = 3 only d)both x = –2 and x = 3Correct answer is option 'A'. Can you explain this answer?
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