Is #f(x)=sinx# concave or convex at #x=-1#?

Question

Isaiah Allen · Accepted Answer

Since #f''(-1)>0#, we see that #sinx# is convex ("concave up") at #x=-1#.

We have to know that #d/dxsinx=cosx# and #d/dxcosx=-sinx#. Also recall that concavity and convexity are determined through the sign of the second derivative of a function. First finding the second derivative: #f(x)=sinx# #f'(x)=cosx# #f''(x)=-sinx# If #f''(-1)>0#, then #f# is convex (commonly called "concave up") at #x=-1#. If #f''(-1)<0#, then #f# is concave (commonly called "concave down") at #x=-1#. We see that #f''(-1)=-sin(-1)approx08415# Since #f''(-1)>0#, we see that #sinx# is convex ("concave up") at #x=-1#.

Adam Adkins · Answer

undefined To determine if $ f(x) = \sin(x) $ is concave or convex at $ x = -1 $, we need to find the second derivative of $ f(x) $ and evaluate it at $ x = -1 $. The second derivative will tell us about the concavity of the function.

The first derivative of $ f(x) = \sin(x) $ is $ f'(x) = \cos(x) $.

The second derivative of $ f(x) $ is $ f''(x) = -\sin(x) $.

When $ x = -1 $, $ f''(-1) = -\sin(-1) $.

Since $ \sin(-1) $ is positive, $ f''(-1) $ is negative.

A function is concave where its second derivative is negative, so at $ x = -1 $, $ f(x) = \sin(x) $ is concave.