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Let f (x) = (4–x 2 ) 2/3 , then f has a (A) a local maxima at x = 0 (B) a local maxima at x = 2 (C) a local maxima at x = –2 (D) none of these

 Let f (x) = (4–x2)2/3, then f has a

 (A) a local maxima at x = 0                   (B) a local maxima at x = 2

 (C) a local maxima at x = –2                 (D) none of these

Grade:12

2 Answers

Arun
25750 Points
4 years ago
Dear student
 
f(x) = (2/3) *(-2x)/(4-x²)-1/3
 
f(x) = -4 x/ 3 (4 -x²)-1/3
 
 
Hence maxima at x = -2 and minima at x = 2
 
Hence option C is correct
Aditya Gupta
2081 Points
4 years ago
arun seems to have totally lost it, as he’s been repeatedly providing WRONG answers for a while now. the correct soln is as follows:
find first derivative
f’(x)= – 4x/[3(4 – x²)1/3]
= (4/3)*x/[{(x – 2)^(1/3)}{(x+2)^(1/3)}]
draw the sign scheme of f’(x), we note that it is negative in interval ( – inf, – 2)U(0, 2) and +ve in ( – 2, 0)U(2, inf). it is zero at x=0. it is not defined at x= ±2.
since f’(x) changes sign from +ve to – ve around x=0 and is zero at x=0, we conclude from the 1st derivative test that f(x) has a local maxima at x= 0. further, you can draw a graph of f(x) to visualise the same.
option (A)
KINDLY APPROVE :))

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