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

#y=sqrt(4x-7)-2/3(x-4)#
#(d^2y)/dx^2=(-4)/((4x-7)sqrt(4x-7)#

1 Answer
Jim G.
Jan 20, 2018

#"maximum at "(4,3)#

Explanation:

#"differentiate using the "color(blue)"chain rule"#

#"given "y=f(g(x))" then"#

#dy/dx=f'(g(x)xxg'(x)larrcolor(blue)"chain rule"#

#y=(4x-7)^(1/2)-2/3(x-4)#

#rArrdy/dx=1/2(4x-7)^(-1/2)xx4-2/3#

#color(white)(rArrdy/dx)=2(4x-7)^(-1/2)-2/3#

#"to obtain the second derivative differentiate "dy/dx#

#rArr(d^2y)/dx^2=-1(4x-7)^(-3/2)xxd/dx(4x-7)#

#color(white)(rArr(d^y)/dx^2)=-4(4x-7)^(-3/2)#

#color(white)(rArr(d^2y)/dx^2)=-4/(4x-7)^(3/2)=-4/((4x-7)sqrt(4x-7))#

#"to find max/min equate "dy/dx=0#

#rArr2(4x-7)^(-1/2)-2/3=0#

#rArr2/(4x-7)^(1/2)=2/3#

#rArr(4x-7)^(1/2)=3#

#rArr4x-7=9rArrx=4#

#"and "y=sqrt(16-7)-2/3(4-4)=3#

#"find the nature of the point using the "color(red)"second derivative test"#

#• " if "(d^2y)/dx^2>0" then minimum"#

#• " if "(d^2y)/dx^2<0" then maximum"#

#x=4to(d^2y)/dx^2=-4/((9)(3))<0#

#rArr(4,3)" is a maximum point"#
graph{sqrt(4x-7)-2/3(x-4) [-10, 10, -5, 5]}

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