Examples Determine the Critical Points for y = 1/3 x3 - 4x2 - 6x + 8. Determine their nature. Sketch the critical graph. Hence Determine the critical points for y = 6x3 + x2 − 4x + 1. Sketch the graph.

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

$$y' = \frac{d}{dx}\left(\frac{1}{3}x^3 + \frac{1}{2}x^2 - 6x + 8\right)$$ $$y' = \frac{1}{3}(3x^2) + \frac{1}{2}(2x) - 6$$ $$y' = x^2 + x - 6$$

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

$$x^2 + x - 6 = 0$$

Factor the quadratic equation:

$$(x+3)(x-2) = 0$$

This gives two critical $x$-values:

$$x = -3 \quad \text{or} \quad x = 2$$

Now, substitute these $x$-values back into the original function $y = \frac{1}{3}x^3 + \frac{1}{2}x^2 - 6x + 8$ to find the corresponding $y$-values.

For $x = -3$:

$$y = \frac{1}{3}(-3)^3 + \frac{1}{2}(-3)^2 - 6(-3) + 8$$ $$y = \frac{1}{3}(-27) + \frac{1}{2}(9) + 18 + 8$$ $$y = -9 + 4.5 + 26$$ $$y = 21.5$$

The first critical point is $(-3, 21.5)$.

For $x = 2$:

$$y = \frac{1}{3}(2)^3 + \frac{1}{2}(2)^2 - 6(2) + 8$$ $$y = \frac{1}{3}(8) + \frac{1}{2}(4) - 12 + 8$$ $$y = \frac{8}{3} + 2 - 4$$ $$y = \frac{8}{3} - 2$$ $$y = \frac{8}{3} - \frac{6}{3} = \frac{2}{3}$$

The second critical point is $(2, \frac{2}{3})$.

The critical points are $\boxed{(-3, 21.5) \text{ and } (2, \frac{2}{3})}$.

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

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

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

For $x = -3$:

$$y''(-3) = 2(-3) + 1 = -6 + 1 = -5$$

Since $y''(-3) < 0$, the critical point $(-3, 21.5)$ is a local maximum.

For $x = 2$:

$$y''(2) = 2(2) + 1 = 4 + 1 = 5$$

Since $y''(2) > 0$, the critical point $(2, \frac{2}{3})$ is a local minimum.

The function is a cubic polynomial with a positive leading coefficient ($\frac{1}{3}x^3$). 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 $(-3, 21.5)$, then decrease until it reaches the local minimum at $(2, \frac{2}{3})$, and finally increase indefinitely.