Is #f(x)=cosx# concave or convex at #x=(3pi)/2#?

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Feb 9, 2018

Answer:

See below.

Explanation:

We can determine where a function is convex or concave, by using the second derivative. If:

#(d^2y)/(dx^2)f(x)>0color(white)(888)# Convex( concave up )

#(d^2y)/(dx^2)f(x)<0color(white)(888)# Concave( concave down )

The second derivative is just the derivative of the first derivative. .i.e.

#(d^2y)/(dx^2)f(x)=dy/dx(dy/dxf(x))#

For #f(x)=cos(x)#

#dy/dx=-sin(x)#

#(d^2y)/(dx^2)(cos(x))=dy/dx(-sin(x))=-cos(x)#

Now we solve the inequalities:

#-cos(x)>0color(white)(888)[1]#

#-cos(x)<0color(white)(888)[2]#

#-cos(x)>0color(white)(888)[1]#

#pi/2 < x< (3pi)/2#

#-cos(x)<0color(white)(888)[2]#

#0 < x < pi/2color(white)(88)# , #(3pi)/2 < x < 2pi #

Notice that #bb((3pi)/2)# is on the point where the function changes from convex to concave. This is called a point of inflection ( inflexion in the UK ), so at #bb((3pi)/2)# it is neither concave nor convex.

This is verified by its graph:

enter image source here

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