How did you figure whether they're maxima or minima?
Find the two stationary points of y=(3x^2+4)/x and determine if they are a local maximum, local minimum or neither.
I've made an attempt at the question and was wondering if someone could please check my answer and guide me if I've done something wrong
y = (3x^2+4)/x
= 3x+(4/x)
=3x+4x^-1
y' = 3-4x^2
0=3-4x^2
x = (-sqrt(3)/2) and (sqrt(3)/2)
Subbing the x into the original equation, y = -25/2sqrt(3) and 25/2sqrt(3)
I got that (-sqrt(3)/2,-25/2sqrt(3)) is a local minimum
and (sqrt(3)/2, 25/2sqrt(3)) is a local maximum.
Am I correct?
Hi Prove it,
I tried finding the gradient on each side using the first derivative:
so when x is -sqrt(3)/2 or approx -0.866, y = 0
when x = -0.87, y = -0.0276 (negative)
when x = -0.85, y = 0.11 (positive) which means if it goes from negative to 0 to positive, then it is a local minimum.
I did the same method with the other stationary point and it went from positive to 0 to negative, meaning it is a local maximum.