## Precalculus: Mathematics for Calculus, 7th Edition

SOLUTION $x^{2}(x-1)(x^{2}+x+1) \gt0$ INTERVAL $(1,\infty)$ GRAPH In the real number line put a one with an open circle with a line pointing towards $\infty$.
$x^{5} \gt x^{2}$ $x^{5}-x^{2} \gt0$ $x^{2}(x^{3}-1) \gt0$ $x^{2}(x-1)(x^{2}+x+1) \gt0$ Now find the real solutions. $x^{2}=0$ $x=0$ $x-1=0$ $x=1$ $x^{2}+x+1$ no real solution INTERVAL $(-\infty, 0) (0,1) (1,\infty)$ $(-\infty, 0)$ $y(-2)=(-2)^{2}(-2-1)((-2)^{2}+(-2)+1) \gt0= y(0)=-36$ No solution $(0,1) Y(0.5)=(0.5)^{2}(0.5-1)(0.5^{2}+0.5+1) \gt0= y(0.5)=-0.219$ No solution $(1,\infty)$Y(3)=$3^{2}(3-1)(3^{2}+3+1) \gt0= y(3)=234$ Solution CORRECT $(1,\infty)$ GRAPH In the real number line put a one with an open circle with a line pointing towards $\infty$.