How do you find all local maximum and minimum points using the second derivative test given #y=4x+sqrt(1-x)#?

Question

Isabella Ackerman · Accepted Answer

#color(blue)("Maximum point at " x=63/64)#

At local maximum and minimum points the tangent is zero. We can use this to find where these points are. We first find the first derivative and equate this to zero.

Given #f(x)=4x+sqrt(1-x)#

#dy/dx(4x+sqrt(1-x))=4x+(1-x)^(1/2)=4+1/2(1-x)^(-1/2)*(-1)#

#=4-1/(2sqrt(1-x))#

Equate to zero:

#4-1/(2sqrt(1-x))=0#

#8sqrt(1-x)-1=0=>x=63/64#

Using the second derivative we know that if:

#(d^2y)/(dx^2)>0color(white)(888)#Minimum point.

#(d^2y)/(dx^2)<0color(white)(888)#Maximum point.

#(d^2y)/(dx^2)=0color(white)(888)#Maximum/Minimum or inflection point.

#(d^2y)/(dx^2)(4x+(1-x)^(1/2))=dy/dx(4-1/(2sqrt(1-x)))#

#dy/dx(4-1/(2sqrt(1-x)))=dy/dx(4-1/2(1-x)^(-1/2))#

#=0+1/4(1-x)^(-3/2)*(-1)=-1/(4(1-x)^(3/2))#

Plugging in #x=63/64#

#-1/(4(1-(63/64))^(3/2))=-4sqrt((1/64)^3)#

#-4sqrt((1/64)^3)<0#

#:.#

#x=63/64# is a maximum point.

This is confirmed by the graph:

Ezra Adams · Answer

undefined To find all local maximum and minimum points using the second derivative test for $ y = 4x + \sqrt{1-x} $, follow these steps:

1. Find the first derivative $ y' $ of the function.
2. Find the second derivative $ y'' $ of the function.
3. Set $ y' = 0 $ and solve for $ x $ to find critical points.
4. Use the second derivative test to determine if the critical points are local maxima, local minima, or neither.