Calculus: Early Transcendentals 8th Edition

There are two inflection points when $c \lt -2\sqrt{\frac{2}{3}}~~$ or $~~c \gt 2\sqrt{\frac{2}{3}}$ There is one inflection point when $c = -2\sqrt{\frac{2}{3}}~~$ or $~~c = 2\sqrt{\frac{2}{3}}$ There are no inflection points when $-2\sqrt{\frac{2}{3}} \lt c \lt 2\sqrt{\frac{2}{3}}$ As $c$ decreases, $P(x)$ does not fall as far below the x-axis, and as $c$ continues to decrease, $P(x)$ does not fall below the x-axis at all.
$P(x) = x^4+cx^3+x^2$ $P'(x) = 4x^3+3cx^2+2x$ $P''(x) = 12x^2+6cx+2$ We can find the values of $x$ such that $P''(x) = 0$: $P''(x) = 12x^2+6cx+2 = 0$ $6x^2+3cx+1 = 0$ We can use the quadratic formula: $x = \frac{-3c\pm \sqrt{(3c)^2-(4)(6)(1)}}{2(6)}$ $x = \frac{-3c\pm \sqrt{9c^2-24}}{12}$ $x = -\frac{c}{4}\pm \frac{\sqrt{9c^2-24}}{12}$ There are two inflection points when $9c^2 \gt 24$: $c^2 \gt \frac{8}{3}$ $c \lt -2\sqrt{\frac{2}{3}}~~$ or $~~c \gt 2\sqrt{\frac{2}{3}}$ There is one inflection point when $9c^2 = 24$: $c^2 = \frac{8}{3}$ $c = -2\sqrt{\frac{2}{3}}~~$ or $~~c = 2\sqrt{\frac{2}{3}}$ There are no inflection points when $9c^2 \lt 24$: $c^2 \lt \frac{8}{3}$ $-2\sqrt{\frac{2}{3}} \lt c \lt 2\sqrt{\frac{2}{3}}$ We can see a sketch of four graphs of $P(x)$ when $c = 1,2,3,4$ As $c$ decreases, $P(x)$ does not fall as far below the x-axis, and as $c$ continues to decrease, $P(x)$ does not fall below the x-axis at all.