## Calculus: Early Transcendentals (2nd Edition)

(a) $r_1 =-\sqrt{\frac{1}{6}(\sqrt{21}-3)}$ and $r_2 =+\sqrt{\frac{1}{6}(\sqrt{21}-3)}.$ are critical points. (b) There is a local minimum at $r_1$ and a local maximum at $r_2$. (c) The absolute maximum is $f(r_2) ≈ .339$ and the absolute minimum is $f(r_1) ≈ −.339$.
(a) . $f(x) = \frac{1}{x^2+1}− 3x^2 = \frac{−3x^4−3x^2+1}{x^2+1}$ . This is $0$ when $−3x^4 − 3x^2 + 1 = 0$. Letting $u = x^2$, we seek roots of $−3u^2 − 3u + 1$. Using the quadratic formula and solving for $u$, and then writing in terms of $x$, we have the roots $x =\pm \sqrt{\frac{1}{6}(\sqrt{21}-3)} ≈ ±.5135$. Let $r_1 =-\sqrt{\frac{1}{6}(\sqrt{21}-3)}$ and $r_2 =+\sqrt{\frac{1}{6}(\sqrt{21}-3)}.$ (b). Note that $f < 0$ on $(−1, r_1)$ and $f > 0$ on $(r_1, r_2)$, and $f < 0$ on $(r_2, 1)$. Thus there is a local minimum at $r_1$ and a local maximum at $r_2$. (c). Note that $f(−1) = −π/4+1 ≈ .215$ and $f(1) = π/4−1 ≈ −.215$, and $f(r_1) ≈ −.339$ and $f(r_2) ≈ .339$. The absolute maximum is $f(r_2) ≈ .339$ and the absolute minimum is $f(r_1) ≈ −.339$.