Can you please identify the min,max & inflection points of function 4x / (x^2+1)?

Question

please give as much detail as possible.

Answer 1

Do it on ur ti89.

Answer 2

You need to take the first derivative to find the min and max, and the second derivative to find the inflection points.

Set the derivatives equal to zero, and then draw a number line. Plug points into the derivative equation on either side of the zeros of the first derivative. If at the zero, the values change from negative to positive, the original curve changes from decreasing to increasing and that is a minimum. If at the zero, the values change from positive to negative, or increasing to decreasing, that is a maximum.

Answer 3

(f(x)/g(x))' = (f'(x)g(x) - f(x)g'(x))/(g(x))^2

(4x/(x^2+1))' = (4(x^2+1) - 4x*2x)/(x^2+1)^2 = 0

4x^2 - 8x^2 +4 = 0

4x^2 = 4

x = 1, -1

Max (1,2) Min(-1, -2)

2nd derivative:

4(1-x^2)/(x^2+1)^2 =

=4(1-x^2)/(x^4+2x^2+1)

" = 4(-2x(x^4+2x^2+1)-(1-x^2)(4x^3+4x))/(x^4+2x^2+1)^2=0

-2x^5-4x^3-2x-4x^3-4x+4x^5+4x^3=0

2x^5-4x^3-6x=0

x(x^4-2x^2-3)=0, x=0 or
let A=x^2
A^2-2A-3=0
(A-3)(A+1)=0 A=3 x=+ - sqt(3)