Gegee: $f(x) = 3x^3$ 9.1 Los op $f(x) = f' (x)$ 9.2 Die grafieke $f$, $f'$ en $f''$ gaan almal deur die punt $(0; 0)$ - NSC Mathematics - Question 9 - 2019 - Paper 1

For the graphs of $f$ and $f'$, the point $(0; 0)$ is a stationary point for $f'$, as $f'(0) = 0$. This means the function $f$ has a turning point or potential inflection point at this coordinate.

The point $(0; 0)$ is identified as an inflection point of $f$ and a turning point of $f'$. This indicates that while $f$ may change concavity there, $f'$ is at a local maximum or minimum, marking different types of stationary behavior.

To find the vertical distance between $f'$ and $f''$ at $x = 1$, we first calculate:

$f'(1) = 9(1)^2 = 9$

And then,

$f''(x) = 18$

Hence, at $x = 1$, the distance is:

$|f'(1) - f''(1)| = |9 - 18| = 9$

To solve $f(x) - f'(x) < 0$, we rewrite:

$3x^3 - 9x^2 < 0$

Factoring out $3x^2$, we have:

$3x^2(x - 3) < 0$

The critical points are $x = 0$ and $x = 3$. Testing intervals reveals:

• For $x < 0$: $3x^2 > 0$, $x - 3 < 0$; product is < 0.
• For $0 < x < 3$: $3x^2 > 0$, $x - 3 < 0$; product is < 0.
• For $x > 3$: $3x^2 > 0$, $x - 3 > 0$; product is > 0.

Thus, the solution is:

$x < 0 \text{ or } 0 < x < 3$