For what values of x is #f(x)=4/x^2+1# concave or convex?

1 Answer
Apr 14, 2016

#f# is convex on the interval #(-oo,0)uu(0,+oo)#.

Explanation:

The determine when a function is concave or convex, analyze the sign, positive or negative, of the function's second derivative:

  • When #f''>0#, then #f# is convex.
  • When #f''<0#, then #f# is concave.

So, we first must find #f''#.

Note that we can write #f# as

#f(x)=4x^-2+1#

Now, through the power rule, we see that

#f'(x)=-8x^-3#

#f''(x)=24x^-4=24/x^4#

We must now determine when #24/x^4# is positive or negative.

It's necessary to note that #x^4# will always be positive, so #24/x^4# will also always be positive. Recall that the domain of #f# excludes #0#, so we know that

#f# is convex on the interval #(-oo,0)uu(0,+oo)#.