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

Question

Paisley Johnson · Accepted Answer

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

We must first determine our function's second derivative.

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))#

Two cases need to be handled by us.

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))#

Ezekiel Alexander · Answer

undefined To find the points of inflection for $ y = \sin(x) + \cos(x) $, we need to find where the second derivative changes sign.

1. First, find the first derivative of $ y $ with respect to $ x $, which is $ y' = \cos(x) - \sin(x) $.

2. Next, find the second derivative of $ y $ with respect to $ x $, which is $ y'' = -\sin(x) - \cos(x) $.

3. Set $ y'' $ equal to zero and solve for $ x $ to find the critical points.

$$ -\sin(x) - \cos(x) = 0 $$

$$ \sin(x) + \cos(x) = 0 $$

4. Solve for $ x $ by applying trigonometric identities or graphical methods.

$$ \sin(x) = -\cos(x) $$

$$ \tan(x) = -1 $$

$$ x = \frac{3\pi}{4} + n\pi, \quad \text{where } n \text{ is an integer}$$

5. Once you have the critical points, test the sign of the second derivative on intervals determined by these critical points to determine where the concavity changes.

6. The points where the concavity changes from concave up to concave down, or vice versa, are the points of inflection.