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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?
Verified Answer
Consider function f(x) =(x2-4)2 where x is a real number. Then the fun...
Most Upvoted Answer
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.
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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