Question

Find the critical points of
(4x-3)/(3x^(1/2))

Answer 1

you should say y = (4x-3)/(3x^(1/2))
or
f(x) = (4x-3)/(3x^(1/2))

dy/dx =[ 3x^(1/2) (4) - (4x-3)(3/2)x^(-1/2) ]/(9x)
= 0 for a max/min

12x^(1/2) = (3/2)(4x-3)/x^(1/2)
12x = (3/2)(4x - 3)
24x = 12x - 9
12x=-9
x = -3/4 , but x ≥0 or else x^(1/2) is undefined.

There are no turning points. (no max/min)

dy/dx can be reduced to (4x-3)/(6x^(3/2))
I will leave it up to you to show that the second derivative is
(9-4x)/(12x^(5/2) )

when we set that equal to zero , we get
x = 9/4
So there is a point of inflection when x = 9/4

what about intercepts?
let x = 0, y is undefined
let y = 0 , x = 3/4

I let Wolfram graph it, and got
http://www.wolframalpha.com/input/?i=y+%3D+(4x-3)%2F(3x%5E(1%2F2))+,+from+0+to+3

I am unable to explain why no point of inflection is shown, but my algebra shows that there is one.

Answer 2

it's because y"=0 when x = -9/4
but y is undefined for x<0.

check for +/- sign mistake