For what values of x is #f(x)=3+x^2/2# concave or convex?

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Answer 1 · 2016-01-01T18:26:42

#f# is convex on #(-oo,+oo)#.

#f# is convex when #f''>0#.
#f# is concave when #f''<0#.

First, find #f'(x)#, then #f''(x)#.

#f(x)=3+1/2x^2#

#f'(x)=x#

#f''(x)=1#

Since #f''(x)# is always #>0#, #f# is convex on #(-oo,+oo)#.

graph{3+(x^2)/2 [-14.96, 17.08, -1.67, 14.35]}

The answer makes logical since, since #f# is a parabola that opens up, so it will always be in the convex shape.