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?

1 Answer
Jan 20, 2018

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