# Find where the maxima and minima of the function f(x)=2x / (x^2+3) occur ?

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We are given the function f(x) = (2*x)/(x^2+3).

We calculate the absolute minimum and and the absolute maximum as follows. First find the differential of the function.

f(x) = (2*x)/(x^2+3).

=> f'(x) = 2/(x^2 +3) - 4x^2 /( x^2 + 3)^2

Now equate this to zero.

2/(x^2 +3) - 4x^2 /( x^2 + 3)^2 = 0

=> 2*( x^2 + 3) - 4x^2 = 0

=> 2x^2 + 6 - 4x^2 =0

=> 2x^2 = 6

=> x^2 = 3

=> x = -sqrt 3 and + sqrt 3

**We have the minima at x= -sqrt 3 and the maxima at x =sqrt 3.**