Question

Given that,`y=sqrt(4x-7)-2/3(x-4)` find `dy/dx` and show that`(d^2y)/dx^2=(-4)/((4x-7)sqrt(4x-7)`. Hence find the maximum point of the curve?

Answer 1

The maximum is at x = 4.

Explanation:

we have been given that
`y = (4x-7)^(1/2) - 2x/3 + 8/3`.
`dy/dx = 1/2 * (4x-7)^(1/2-1)* 4 - 2/3`
`=> dy/dx = 2*(4x-7)^(-1/2) -2/3`

now ,
`(d^2y)/dx^2 = 2*-1/2*(4x-7)^(-1/2 -1)*4`

=>`(d^2y)/dx^2 = -4(4x-7)^-(3/2)`

=>`(d^2y)/dx^2 = -4/((4x-7)*sqrt(4x-7))`. hence proved

now to find maximum point ,
we know that`dy/dx = 0` at maximum and minimum points. So,
`dy/dx = 2/sqrt(4x-7)-2/3 = 0`

=>`sqrt(4x-7) = 3`
=> `4x-7 = 9` (squaring on both sides)
`4x = 9+7 = 16`
and therefore `x = 16/4 =4.`

remember at this point the function could be max or minimum.
In order to say that at x = 4 is maximum , do second derivative test

`(d^2y)/dx^2 = -4/((4*4-7)*sqrt(4*4-7)) = -4/(9*3) = -4/27`
which is less than 0 . So x=4 is the maximum point of the curve