Sketched below is the graph of $f(x) = x^{3} + ax^{2} + bx + c$ - NSC Mathematics - Question 9 - 2022 - Paper 1

To find the coordinates of the turning point $N$, we need to find the first derivative of $f(x)$, set it to zero, and solve for $x$:

1. The first derivative is:
   $f'(x) = 3x^{2} - 2x - 5$.

2. Set $f'(x) = 0$:
   $3x^{2} - 2x - 5 = 0$.

   Using the quadratic formula x = rac{-b ext{± } ext{√}(b^{2}-4ac)}{2a}:

   Plugging in $a=3$, $b=-2$, and $c=-5$ gives:

   x = rac{2 ext{± } ext{√}(4 + 60)}{6} = rac{2 ext{± } ext{√}(64)}{6} = rac{2 ext{± } 8}{6}.

   Therefore, x_n = rac{10}{6} = rac{5}{3} or x_n = rac{-6}{6} = -1.

3. Substitute back into the function f(x) to find $y$:
   Better to choose x = rac{5}{3} or $x=-1$ to find $N$ coordinates so we use coordinate pair:

   Nigg( rac{5}{3}, figg( rac{5}{3} igg) igg) = N igg(rac{5}{3}, -rac{67}{27} igg).

To determine when $f(x) < 0$, we need to analyze the sign of $f(x)$ over its intervals:

1. Utilizing the roots we earlier determined, we note that the intervals to test are:

   • $(- ext{∞}, -1)$,
   • $(-1, 3)$,
   • and $(3, ext{∞})$.

2. Testing a value within each interval, for example:

   • For $x = -2$:
     f(-2) = $(-2)^{3} + (-2)^{2} - 5(-2) - 3 = -8 + 4 + 10 - 3 = 3 > 0$.
   • For $x = 0$:
     f(0) = $-3 < 0$.
   • For $x = 4$:
     f(4) = $64 - 16 - 20 - 3 = 25 > 0$.

3. Therefore, $f(x) < 0$ for $x ext{ in } (-1, 3)$.

To find where $f$ is increasing, we examine the sign of $f^{ ext{'}}(x)$:

1. We know $f^{ ext{'}}(x) = 3x^{2} - 2x - 5$ and set this greater than zero:

   $3x^{2} - 2x - 5 > 0$.

2. Based on our solutions for earlier turning points, we established intervals:

   • $(- ext{∞}, -1)$: Test a point such as $-2$ gives > 0.
   • $(-1, 5/3)$: Test a value of $0$, hence gives negative.
   • $(5/3, ext{∞})$: Test say $4$ hence gives positive > 0.

3. So, $f$ is increasing on $(- ext{∞}, -1)$ and $(5/3, ext{∞})$.

To determine the concavity of $f$, we require the second derivative:

1. Compute the second derivative:
   $f^{ ext{''}}(x) = 6x - 2$.

2. Setting $f^{ ext{''}}(x) > 0$ gives:

   6x - 2 > 0 ightarrow x > rac{1}{3}.

3. Therefore, $f$ is concave up for all $x$ values where x > rac{1}{3}.

To find the distance between the graphs of $f(x)$ and $f^{ ext{'}}(x)$:

1. Write the distance function:
   $d(x) = |f(x) - f^{ ext{'}}(x)|$ where $x$ is in the interval.

2. Compute $d(x)$:

   From earlier findings,

   $f(x) = x^{3}-x^{2}-5x-3$ and $f^{ ext{'}}(x) = 3x^{2}-2x-5$.
   $d(x) = |(x^{3}-x^{2}-5x-3) - (3x^{2}-2x-5)|$ $= |x^{3} - 4x^{2} -3 + 2| = |x^{3} - 4x^{2} + 2|$

3. Find local maxima of this distance function by evaluating critical points in that interval, checking the values at $-1$ and values approaching $0$ (evaluate if differentiable).

4. Finally, compute the maximum vertical distance and employ the maximum identity to find it over the computed interval values, applying limits as necessary.