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

Question

Ezra Adams · Accepted Answer

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

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

Ezra Adams · Answer

undefined To determine the concavity of $ f(x) = \frac{4}{x^2 + 1} $, we need to find the second derivative of $ f(x) $ and then analyze its sign. Let $ f''(x) $ represent the second derivative of $ f(x) $.

$ f''(x) = \frac{d^2}{dx^2} \left( \frac{4}{x^2 + 1} \right) = \frac{d}{dx} \left( \frac{-8x}{(x^2 + 1)^2} \right) = \frac{-8(x^2 - 1)}{(x^2 + 1)^3} $

Now, to determine the concavity or convexity, we analyze the sign of $ f''(x) $.

1. $ f''(x) > 0 $ for $ x \in (-\infty, -1) \cup (1, \infty) $, meaning $ f(x) $ is concave up on these intervals.
2. $ f''(x) < 0 $ for $ x \in (-1, 1) $, meaning $ f(x) $ is concave down on this interval.