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