Given: $f(x) = x^3 - x^2 - x + 1$ 9.1 Write down the coordinates of the $y$-intercept of $f$ - NSC Mathematics - Question 9 - 2017 - Paper 1

To find the $x$-intercepts, set $f(x)$ to 0:

$x^3 - x^2 - x + 1 = 0$

This can be factored as:

$(x-1)(x^2 + 1) = 0$

The solutions are:

1. $x - 1 = 0 ightarrow x = 1$
2. $x^2 + 1 = 0$ has no real solutions.

Thus, the $x$-intercepts are $(-1, 0)$ and $(1, 0)$.

The turning points can be found by finding the derivative:

$f'(x) = 3x^2 - 2x - 1$

Setting the derivative to 0:

$3x^2 - 2x - 1 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6}$

Thus, $x = 1$ or $x = -\frac{1}{3}$.

Next, substituting these values back into $f(x)$ gives:

1. For $x = 1$: $f(1) = 1 - 1 - 1 + 1 = 0$ Therefore, turning point is $(1, 0)$.
2. For $x = -\frac{1}{3}$: $f\left(-\frac{1}{3}\right) = -\frac{1}{27} - \frac{1}{9} + \frac{1}{3} + 1 = \left(-\frac{1 + 3 - 9}{27}\right) = \left(-\frac{1}{3}, \frac{32}{27}\right)$ Thus, $( -\frac{1}{3}, \frac{32}{27})$ is another turning point.

Ensure to mark the following points on your sketch:

• $y$-intercept: $(0, 1)$
• $x$-intercepts: $(-1, 0)$ and $(1, 0)$
• Turning points: $(1, 0)$ and $(-\frac{1}{3}, \frac{32}{27})$.

The general shape of the graph should reflect these intercepts and turning points, ensuring smooth curves between the identified points.

From the roots found in the derivative, analyze the intervals:

1. For $x < -\frac{1}{3}$: f^{rown}(x) is positive.
2. For $-\frac{1}{3} < x < 1$: f^{rown}(x) is negative.
3. For $x > 1$: f^{rown}(x) is positive.

Thus, the answer is: $-\frac{1}{3} < x < 1$.