Quick Calculus Question about Extrema?

Question

Ok i am stuck on this question.....

Locate and Classify all extrema for .....

f(x) = 6 / (x^2 + 3) on [-3,1]

Answer 1

Take first derivative and set to zero

df/dx = -12x/(x^2+3)^2 = 0 ---> x = 0 is an extrema

take second derivative and evaluate at x = 0

d^2f/dx^2 = -12/(x^2+3)^2 +48x^2/(x^2+3)^3 --> -12/9 at x = 0

So since d^2/dx^2 < 0, the point x = 0 is a maximum.

Answer 2

f(x) = 6 / (x^2 + 3) on [-3,1]
when x = 0 . . . . then y = 2 . . .. this is the highest point
asymptote .. . . y = 0
(-3, 1) . . . . . does not touch lie on the curve