Question

sketch the curve x^3-x^2-6x

Answer 1

Answer on Question #50547 – Math - Differential Calculus | Equations

sketch the curve $x^3 - x^2 - 6x$

Solution.

$$y = x^3 - x^2 - 6x$$

1) Domain: $-\infty < x < \infty$.

2) Range: $-\infty < y < \infty$. Note that $y \to \infty$ as $x \to \infty$, $y \to -\infty$ as $x \to -\infty$.

3) Intercepts:

$$\begin{array}{l} x \text{ - intercepts: } y = 0 \to x^3 - x^2 - 6x = 0 \to x(x^2 - x - 6) = 0 \to \\ x(x - 3)(x + 2) = 0 \to x = 0, x = 3, x = -2, \text{ hence } (0, 0), \\ (-2, 0), (3, 0) \text{ are } x \text{ - intercepts} \end{array}$$ $$y \text{ - intercept: } x = 0 \to y(0) = 0^3 - 0^2 - 6 \cdot 0 = 0, \text{ hence}$$

$(0, 0)$ is $y$ - intercept.

4) Symmetry: $y(x)$ is neither even nor odd function,

$$\text{because } y(-x) \neq -y(x), y(-x) \neq y(x)$$

5) Asymptotes: there are no asymptotes.

6) Local extrema: $y' = 0 \to 3x^2 - 2x - 6 = 0 \to x = \frac{1 - \sqrt{19}}{3}$, or $x = \frac{1 + \sqrt{19}}{3}$.

$$\left(\frac{1 - \sqrt{19}}{3}, \frac{38\sqrt{19} - 56}{27}\right) \text{ - local maximum};$$ $$\left(\frac{1 + \sqrt{19}}{3}, \frac{-38\sqrt{19} - 56}{27}\right) \text{ - local minimum};$$

7) Intervals of increase and decrease:

$$\text{increasing on } \left(-\infty, \frac{1 - \sqrt{19}}{3}\right) \text{ and on } \left(\frac{1 + \sqrt{19}}{3}, \infty\right);$$

decreasing on $\left(\frac{1 - \sqrt{19}}{3},\frac{1 + \sqrt{19}}{3}\right)$

8) Concavity and points of inflection:

$$y ^ {\prime \prime} (x) = 0 \rightarrow 6 x - 2 = 0 \rightarrow x = \frac {1}{3};$$

concave downward on $\left(-\infty ,\frac{1}{3}\right)$ , concave upward on $\left(\frac{1}{3},\infty\right)$ ;

point of inflection is $\left(\frac{1}{3}, -\frac{19}{3}\right)$ .

Graph of $y$ as function of $x$ is shown below:

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