How do you find points of inflection for #y= sin x + cos x#?

1 Answer
Jul 14, 2015

The point of inflexion are: #((3pi)/4+2kpi , 0) " AND "((-pi/2+2kpi,0))#

Explanation:

1 - First we have to find the second derivative of our function.

2 - Second, we equate that derivative#((d^2y)/(dx^2))# to zero

#y=sinx +cosx#

#=>(dy)/(dx)=cosx-sinx#

#=>(d^2y)/(dx^2)=-sinx-cosx#

Next, #-sinx-cosx=0#

#=>sinx+cosx=0#

Now, we shall express that in the form #Rcos(x+lamda)#

Where #lambda# is just an acute angle and #R# is a positive integer to be determined. Like this

#sinx+cosx=Rcos(x+lambda)#

#=>sinx+cosx=Rcosxcoslamda - sinxsinlamda#

By equating the coefficients of #sinx# and #cosx# on either side of the equation,

#=>Rcoslamda=1#

and #Rsinlambda=-1#

#(Rsinlambda)/(Rcoslambda)=(-1)/1=>tanlambda=-1=>lambda=tan^-1(-1)=-pi/4#

And #(Rcoslambda)^2+(Rsinlambda)^2=(1)^2+(-1)^2#

#=>R^2(cos^2x+sin^2x)=2#

But we know the identity , #cos^2x+sin^2=1#

Hence, #R^2(1)=2=>R=sqrt(2)#

In a nut shell, #(d^2y)/(dx^2)=-sinx-cosx=sqrt(2)cos(x-pi/4)=0#

#=>sqrt(2)cos(x-pi/4)=0#

#=>cos(x-pi/4)=0=cos(pi/2)#

So the general solution of #x# is : #x-pi/4=+-pi/2+2kpi# , #kinZZ#

#=>x=pi/4+-pi/2+2kpi#

So the points of inflexion will be any point that has coordinates :
#(pi/4+-pi/2+2kpi , sqrt(2)cos(pi/4+-pi/2-pi/4))#

We have two cases to deall with,

Case 1

#(pi/4+pi/2+2kpi , sqrt(2)cos(pi/4+pi/2-pi/4))#

#=>((3pi)/4+2kpi , sqrt(2)cos(pi/2))#

#=>((3pi)/4+2kpi , 0)#

Case 2

#(pi/4-pi/2+2kpi , sqrt(2)cos(pi/4-pi/2-pi/4))#

#=>(-pi/2+2kpi , sqrt(2)cos(-pi/2))#

#=>((-pi/2+2kpi , 0))#