JEE Exam > JEE Questions > Let f (x) =(x2−4)1/3, then f has aa)loc...
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Let f (x) = (x2−4)1/3, then f has a
- a)local minima at x = 0
- b)local maxima at x = 0
- c)point of inflextion at x = 0
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Let f (x) =(x2−4)1/3, then f has aa)local minima at x = 0b)local...
Also, for x < 0 (slightly) , f ‘(x) < 0 and for x > 0 (slightly) f ‘(x) >0. Hence f has a local minima at x = 0 .
Most Upvoted Answer
Let f (x) =(x2−4)1/3, then f has aa)local minima at x = 0b)local...
Solution:
To find the local minima or maxima, we need to find the first derivative of the given function and equate it to zero.
Let's find the first derivative of the given function f(x):
f(x) = (x^2/4)^(1/3)
f'(x) = (1/3)(x^2/4)^(-2/3)(2x/4)
f'(x) = (1/6)(x^2)^(1/3)(1/x)
f'(x) = (1/6)(x^2)^(1/3)/x
Now, equating f'(x) to zero:
(1/6)(x^2)^(1/3)/x = 0
(x^2)^(1/3) = 0
x = 0
Therefore, the only critical point of f(x) is x=0.
Now, let's check the nature of this critical point. For this, we need to find the second derivative of f(x) and evaluate it at x=0.
f''(x) = (1/6)(x^2)^(1/3)(-2x/(3x^2)) + (1/6)(1/x)(1/3)(x^2)^(1/3)(2x)
f''(x) = -(1/9)(x^2)^(-1/3) + (1/9)(x^2)^(1/3)
f''(x) = (1/9)(x^2)^(1/3)[1 - (1/x^2)]
f''(0) = (1/9)(0)^(1/3)[1 - (1/0^2)]
f''(0) is undefined.
Since the second derivative of f(x) is undefined at x=0, we cannot determine the nature of this critical point using the second derivative test.
However, we can observe the behavior of f(x) around x=0 to determine its nature.
For x>0, f(x) is an increasing function since its first derivative is positive.
For x<0, f(x)="" is="" a="" decreasing="" function="" since="" its="" first="" derivative="" is="">
Therefore, x=0 is a local minima of f(x).
Hence, the correct answer is option 'A'.
To find the local minima or maxima, we need to find the first derivative of the given function and equate it to zero.
Let's find the first derivative of the given function f(x):
f(x) = (x^2/4)^(1/3)
f'(x) = (1/3)(x^2/4)^(-2/3)(2x/4)
f'(x) = (1/6)(x^2)^(1/3)(1/x)
f'(x) = (1/6)(x^2)^(1/3)/x
Now, equating f'(x) to zero:
(1/6)(x^2)^(1/3)/x = 0
(x^2)^(1/3) = 0
x = 0
Therefore, the only critical point of f(x) is x=0.
Now, let's check the nature of this critical point. For this, we need to find the second derivative of f(x) and evaluate it at x=0.
f''(x) = (1/6)(x^2)^(1/3)(-2x/(3x^2)) + (1/6)(1/x)(1/3)(x^2)^(1/3)(2x)
f''(x) = -(1/9)(x^2)^(-1/3) + (1/9)(x^2)^(1/3)
f''(x) = (1/9)(x^2)^(1/3)[1 - (1/x^2)]
f''(0) = (1/9)(0)^(1/3)[1 - (1/0^2)]
f''(0) is undefined.
Since the second derivative of f(x) is undefined at x=0, we cannot determine the nature of this critical point using the second derivative test.
However, we can observe the behavior of f(x) around x=0 to determine its nature.
For x>0, f(x) is an increasing function since its first derivative is positive.
For x<0, f(x)="" is="" a="" decreasing="" function="" since="" its="" first="" derivative="" is="">
Therefore, x=0 is a local minima of f(x).
Hence, the correct answer is option 'A'.
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