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

Answer 1

#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)#.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer 2

Ezra Adams

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 $f''(x)$ .

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

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.