8. (8 points) identify the x - intercepts and the points where the local maximums and local minimums occur…

8. (8 points) identify the x - intercepts and the points where the local maximums and local minimums occur. determine the intervals for which the function is increasing or decreasing. g(x)=x^{3}-2x^{2}-x + 1 x - intercepts: - 1,1,4 maximum: x = 0 minimum: x = 1 increasing: (-\\infty,0)(1,\\infty) decreasing: (0,1)

8. (8 points) identify the x - intercepts and the points where the local maximums and local minimums occur. determine the intervals for which the function is increasing or decreasing. g(x)=x^{3}-2x^{2}-x + 1 x - intercepts: - 1,1,4 maximum: x = 0 minimum: x = 1 increasing: (-\\infty,0)(1,\\infty) decreasing: (0,1)

Answer

Explanation:

Step1: Find the derivative

Differentiate $g(x)=x^{3}-2x^{2}-x + 1$ using the power - rule. $g'(x)=3x^{2}-4x - 1$.

Step2: Find critical points

Set $g'(x)=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $3x^{2}-4x - 1 = 0$ where $a = 3$, $b=-4$, $c=-1$. $x=\frac{4\pm\sqrt{16+12}}{6}=\frac{4\pm\sqrt{28}}{6}=\frac{4\pm2\sqrt{7}}{6}=\frac{2\pm\sqrt{7}}{3}\approx1.55,-0.22$.

Step3: Find x - intercepts

Set $g(x)=0$. By the rational - root theorem, possible rational roots are $\pm1$. $g(1)=1 - 2-1 + 1=-1$, $g(-1)=-1 - 2 + 1+1=-1$. Using a numerical method (e.g., Newton - Raphson), we find the roots approximately.

Step4: Determine intervals of increase and decrease

Test intervals using $g'(x)$. Choose test points in the intervals $(-\infty,\frac{2 - \sqrt{7}}{3})$, $(\frac{2 - \sqrt{7}}{3},\frac{2+\sqrt{7}}{3})$, $(\frac{2+\sqrt{7}}{3},\infty)$. If $g'(x)>0$, the function is increasing; if $g'(x)<0$, the function is decreasing.

Step5: Find local maxima and minima

Evaluate $g(x)$ at the critical points. The local maximum occurs at $x=\frac{2 - \sqrt{7}}{3}$ and the local minimum occurs at $x=\frac{2+\sqrt{7}}{3}$.

The x - intercepts are found by solving $x^{3}-2x^{2}-x + 1 = 0$. The critical points from $g'(x)=3x^{2}-4x - 1 = 0$ give the local maxima and minima and the intervals of increase and decrease.

The correct values:

  • x - intercepts: By using a numerical method (such as Newton - Raphson method), the x - intercepts of $g(x)=x^{3}-2x^{2}-x + 1$ are approximately $x\approx - 0.61,0.54,2.07$.
  • Local maximum: $x=\frac{2-\sqrt{7}}{3}\approx - 0.22$.
  • Local minimum: $x=\frac{2+\sqrt{7}}{3}\approx1.55$.
  • Increasing intervals: $(-\infty,\frac{2 - \sqrt{7}}{3})\cup(\frac{2+\sqrt{7}}{3},\infty)$.
  • Decreasing interval: $(\frac{2 - \sqrt{7}}{3},\frac{2+\sqrt{7}}{3})$.

Answer:

x - intercepts: approximately $x\approx - 0.61,0.54,2.07$; Local maximum at $x=\frac{2-\sqrt{7}}{3}\approx - 0.22$; Local minimum at $x=\frac{2+\sqrt{7}}{3}\approx1.55$; Increasing intervals: $(-\infty,\frac{2 - \sqrt{7}}{3})\cup(\frac{2+\sqrt{7}}{3},\infty)$; Decreasing interval: $(\frac{2 - \sqrt{7}}{3},\frac{2+\sqrt{7}}{3})$