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Consider function f(x) =(x2-4)2 where x is a real number. Then the function has  
  • a)
    Only one minimum
  • b)
    Only two minima
  • c)
    Three minima
  • d)
    Three maxima 
Correct answer is option 'B'. Can you explain this answer?
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Consider function f(x) =(x2-4)2 where x is a real number. Then the fun...
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Consider function f(x) =(x2-4)2 where x is a real number. Then the fun...
Explanation:
The given function is f(x) = (x^2 - 4)^2.

First Derivative Test:
To find the minima and maxima of the function, we need to find the first derivative of the function.

f'(x) = 4x(x^2 - 4) = 4x(x + 2)(x - 2)

Now, we need to find the critical points by setting f'(x) = 0.

4x(x + 2)(x - 2) = 0

This gives us three critical points: x = -2, x = 0, and x = 2.

Second Derivative Test:
Now, we need to find the nature of these critical points, whether they are maxima or minima.

For this, we need to find the second derivative of the function.

f''(x) = 12x^2 - 24

At x = -2, f''(-2) = 48 > 0, which means that x = -2 is a local minimum.

At x = 0, f''(0) = -24 < 0,="" which="" means="" that="" x="0" is="" a="" local="" />

At x = 2, f''(2) = 48 > 0, which means that x = 2 is a local minimum.

Therefore, the function has only two minima and one maximum.

Answer:
The correct answer is option B, which says that the function has only two minima.
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