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

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

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 1^st 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.