Determine the Critical points for y = 1/3 x3 + 1/2 x2 - 6x + 8. (i) Determine their nature (ii) Hence sketch the graph. 2. Determine the critical points for y = 6x3 + x2 − 4x + 1. Determine their nature. Hence sketch the graph.

To find the critical points, we first calculate the first derivative of the function $y$ with respect to $x$.

$$y' = \frac{d}{dx}(6x^3 + x^2 - 4x + 1)$$ $$y' = 6(3x^2) + 2x - 4$$ $$y' = 18x^2 + 2x - 4$$

Set the first derivative equal to zero to find the $x$-coordinates of the critical points.

$$18x^2 + 2x - 4 = 0$$

Divide the entire equation by 2 to simplify:

$$9x^2 + x - 2 = 0$$

This is a quadratic equation. We can use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=9$, $b=1$, $c=-2$.

$$x = \frac{-1 \pm \sqrt{1^2 - 4(9)(-2)}}{2(9)}$$ $$x = \frac{-1 \pm \sqrt{1 + 72}}{18}$$ $$x = \frac{-1 \pm \sqrt{73}}{18}$$

This gives two critical $x$-values:

$$x_1 = \frac{-1 + \sqrt{73}}{18} \approx \frac{-1 + 8.544}{18} \approx \frac{7.544}{18} \approx 0.419$$ $$x_2 = \frac{-1 - \sqrt{73}}{18} \approx \frac{-1 - 8.544}{18} \approx \frac{-9.544}{18} \approx -0.530$$

Now, substitute these $x$-values back into the original function $y = 6x^3 + x^2 - 4x + 1$ to find the corresponding $y$-values.

For $x_1 = \frac{-1 + \sqrt{73}}{18} \approx 0.419$:

$$y_1 = 6\left(\frac{-1 + \sqrt{73}}{18}\right)^3 + \left(\frac{-1 + \sqrt{73}}{18}\right)^2 - 4\left(\frac{-1 + \sqrt{73}}{18}\right) + 1$$ $$y_1 \approx 6(0.419)^3 + (0.419)^2 - 4(0.419) + 1$$ $$y_1 \approx 6(0.0736) + 0.1756 - 1.676 + 1$$ $$y_1 \approx 0.4416 + 0.1756 - 1.676 + 1 \approx -0.0588$$

The first critical point is approximately $(0.419, -0.059)$.

For $x_2 = \frac{-1 - \sqrt{73}}{18} \approx -0.530$:

$$y_2 = 6\left(\frac{-1 - \sqrt{73}}{18}\right)^3 + \left(\frac{-1 - \sqrt{73}}{18}\right)^2 - 4\left(\frac{-1 - \sqrt{73}}{18}\right) + 1$$ $$y_2 \approx 6(-0.530)^3 + (-0.530)^2 - 4(-0.530) + 1$$ $$y_2 \approx 6(-0.1488) + 0.2809 + 2.12 + 1$$ $$y_2 \approx -0.8928 + 0.2809 + 2.12 + 1 \approx 2.5081$$

The second critical point is approximately $(-0.530, 2.508)$.

The critical points are $\boxed{\left(\frac{-1 + \sqrt{73}}{18}, -0.059\right) \text{ and } \left(\frac{-1 - \sqrt{73}}{18}, 2.508\right)}$.

To determine if these points are local maxima or minima, we use the second derivative test. First, find the second derivative:

$$y'' = \frac{d}{dx}(18x^2 + 2x - 4)$$ $$y'' = 36x + 2$$

Now, evaluate $y''$ at each critical $x$-value:

For $x_1 = \frac{-1 + \sqrt{73}}{18} \approx 0.419$:

$$y''\left(\frac{-1 + \sqrt{73}}{18}\right) = 36\left(\frac{-1 + \sqrt{73}}{18}\right) + 2$$ $$y''\left(\frac{-1 + \sqrt{73}}{18}\right) = 2(-1 + \sqrt{73}) + 2$$ $$y''\left(\frac{-1 + \sqrt{73}}{18}\right) = -2 + 2\sqrt{73} + 2 = 2\sqrt{73}$$

Since $2\sqrt{73} > 0$, the critical point $\left(\frac{-1 + \sqrt{73}}{18}, -0.059\right)$ is a local minimum.

For $x_2 = \frac{-1 - \sqrt{73}}{18} \approx -0.530$:

$$y''\left(\frac{-1 - \sqrt{73}}{18}\right) = 36\left(\frac{-1 - \sqrt{73}}{18}\right) + 2$$ $$y''\left(\frac{-1 - \sqrt{73}}{18}\right) = 2(-1 - \sqrt{73}) + 2$$ $$y''\left(\frac{-1 - \sqrt{73}}{18}\right) = -2 - 2\sqrt{73} + 2 = -2\sqrt{73}$$

Since $-2\sqrt{73} < 0$, the critical point $\left(\frac{-1 - \sqrt{73}}{18}, 2.508\right)$ is a local maximum.

The function is a cubic polynomial with a positive leading coefficient ($6x^3$). This means the graph will generally rise from left to right.

• As $x \to -\infty$, $y \to -\infty$.
• As $x \to \infty$, $y \to \infty$. The graph will increase until it reaches the local maximum at approximately $(-0.530, 2.508)$, then decrease until it reaches the local minimum at approximately $(0.419, -0.059)$, and finally increase indefinitely.