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

The x-intercepts occur when $f(x) = 0$. We solve the equation:

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

By factoring, we can rewrite it as:

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

From this, we find:

• $x + 1 = 0 \Rightarrow x = -1$
• $x^2 - 2x + 1 = 0 \Rightarrow (x-1)^2 = 0 \Rightarrow x = 1$

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

To find the turning points, we first calculate the first derivative:

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

Setting the first derivative equal to zero to find critical points:

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

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find:

$x = \frac{2 \pm \sqrt{(-2)^2 - 4 * 3 * (-1)}}{2 * 3} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm 4}{6}$

Calculating the two roots gives:

• $x_1 = 1$
• $x_2 = -\frac{1}{3}$

Next, we evaluate $f(x)$ at these points:

• For $x = 1$:
  $f(1) = 1^3 - 1^2 - 1 + 1 = 0$
  So, one turning point is $(1, 0)$.
• For $x = -\frac{1}{3}$:
  $f(-\frac{1}{3}) = (-\frac{1}{3})^3 - (-\frac{1}{3})^2 - (-\frac{1}{3}) + 1 = -\frac{1}{27} - \frac{1}{9} + \frac{1}{3} + 1$
  The calculation yields: $f(-\frac{1}{3}) = 1 - 0.037 - 0.111 + 0.333 = \frac{32}{27}$
  Hence, the coordinates of the turning points are $\left(-\frac{1}{3}, \frac{32}{27}\right)$ and $(1, 0)$.

In your sketch:

• Indicate the y-intercept at $(0, 1)$.
• Indicate the x-intercepts at $(-1, 0)$ and $(1, 0)$.
• Plot the turning points at $\left(-\frac{1}{3}, \frac{32}{27}\right)$ and $(1, 0)$.

Ensure the curve reflects its behavior around these points, showing the cubic nature of the function.

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

To find where this derivative is less than zero, we first find the roots:

• Calculated previously as $x = -1$ and $x = \frac{1}{3}$.

Now we analyze the intervals:

1. For $x < -1$, $f'(x) > 0$.
2. For $-1 < x < \frac{1}{3}$, $f'(x) < 0$.
3. For $x > \frac{1}{3}$, $f'(x) > 0$.

Thus, the values of $x$ for which $f'(x) < 0$ are: $\left(-1, \frac{1}{3}\right)$.