91Ó°ÊÓ

To find the intervals of concavity, we first need to find the first and second derivatives of the function \(f(x)\). Let's begin by finding the first derivative, \(f'(x)\), using the quotient rule: $$\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$ Here, \(u = 1\) and \(v = 1 + x^2\). $$f'(x)=\frac{d}{dx} \left(\frac{1}{1+x^{2}}\right) = \frac{(1+x^2)\frac{d}{dx}(1) - 1\frac{d}{dx}(1+x^2)}{(1+x^2)^2}$$ Now, we find the derivatives of \(u\) and \(v\): $$\frac{d}{dx}(1) = 0$$ $$\frac{d}{dx}(1+x^2) = 2x$$ So, the first derivative of the function is: $$f'(x) = \frac{(1+x^2)(0) - 1(2x)}{(1+x^2)^2}= -\frac{2x}{(1+x^2)^2}$$

Now, we'll find the second derivative, \(f''(x)\), using the quotient rule again. Here, \(u = -2x\) and \(v = (1 + x^2)^2\): $$f''(x)=\frac{d}{dx} \left(-\frac{2x}{(1+x^{2})^2}\right) = \frac{(1+x^2)^2\frac{d}{dx}(-2x) - (-2x)\frac{d}{dx}((1+x^2)^2)}{((1+x^2)^2)^2}$$ Now, we find the derivatives of \(u\) and \(v\): $$\frac{d}{dx}(-2x) = -2$$ $$\frac{d}{dx}((1+x^2)^2) = 2(1+x^2)(2x)$$ So, the second derivative of the function is: $$f''(x) = \frac{(1+x^2)^2(-2) - (-2x)(2(1+x^2)(2x))}{((1+x^2)^2)^2} = \frac{2(3x^2 -1)}{(1+x^2)^3}$$

To find the intervals where the function is concave up or down, we'll analyze the sign of \(f''(x) = \frac{2(3x^2 -1)}{(1+x^2)^3}\). The denominator is always positive for all values of \(x\). The sign of the numerator depends on the value of \(3x^2 - 1\). - If \(3x^2 - 1 > 0\), then \(f''(x) > 0\) and the function is concave up. - If \(3x^2 - 1 < 0\), then \(f''(x) < 0\) and the function is concave down. We can solve \(3x^2 - 1 = 0\), and we get \(x = \pm\frac{1}{\sqrt{3}}\). So, - Concave up: \(x < -\frac{1}{\sqrt{3}}\) or \(x > \frac{1}{\sqrt{3}}\) - Concave down: \(-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}\)

Inflection points occur where the second derivative changes its sign. In this case, the inflection points are at \(x = \pm\frac{1}{\sqrt{3}}\). To find the corresponding points on the function, we plug these values into the original function: $$f\left(-\frac{1}{\sqrt{3}}\right) = \frac{1}{1+\left(-\frac{1}{\sqrt{3}}\right)^{2}} = \frac{3}{4}$$ $$f\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}} = \frac{3}{4}$$ So the inflection points are \(\left(-\frac{1}{\sqrt{3}}, \frac{3}{4}\right)\) and \(\left(\frac{1}{\sqrt{3}}, \frac{3}{4}\right)\). In summary, the function is concave up where \(x < -\frac{1}{\sqrt{3}}\) or \(x > \frac{1}{\sqrt{3}}\), concave down where \(-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}\), and has inflection points at \(\left(-\frac{1}{\sqrt{3}}, \frac{3}{4}\right)\) and \(\left(\frac{1}{\sqrt{3}}, \frac{3}{4}\right)\).