Dear friends, Please read our latest blog post for an important announcement about the website. ❤, The Socratic Team

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

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

1
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

Was this helpful? Let the contributor know!
1500
Trending questions
Impact of this question
12 views around the world
You can reuse this answer
Creative Commons License