The sketch below represents the graphs of functions g and f defined by g(x) = 9x + 18 and f(x) = x^3 + bx^2 + cx + d respectively - NSC Technical Mathematics - Question 7 - 2019 - Paper 1

To show the values of b, c, and d, we refer to the function f(x).

1. Substituting known values:

   • We have f(2) and f(3) given:

     $f(2) = 2b + c - 5 = 0 \quad ... (1)$ $f(3) = 3b + c - 15 = 0 \quad ... (2)$

   • From (1): $2b + c - 5 = 0 \Rightarrow c = 5 - 2b \quad (3)$

2. Substituting into (2):

   • Now substitute (3) into (2):

     $3b + (5 - 2b) - 15 = 0$(simplifying gives)

     $b - 10 = 0 \Rightarrow b = 10$

3. Finding c and d:

   • Substitute b back into (3):

     $c = 5 - 2(-4) = 13$

   • Now from additional information, we find d:

     $d = 18$

To find the coordinates of R, we need to evaluate f'(x) and find where it equals zero, indicating turning points.

1. Calculating f'(x):

   $f(x) = x^3 - 4x^2 - 3x + 18$

   Therefore, the derivative is: $f'(x) = 3x^2 - 8x - 3$

2. Setting f'(x) = 0 to find R:

   Use the quadratic formula:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   Here, a = 3, b = -8, c = -3,

   $x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3}$

   Calculate the discriminant and solve for x:

   Upon simplification, this results in two x-values, leading us to the coordinates of R.

The equation of a tangent line at a point can be written as:

$y - f(R) = f'(R)(x - R)$

Calculate f(R) and f'(R) to plug into the equation:

Analyze g(x):

To determine where g(x) is positive, we find where:

$g(x) = 9x + 18 > 0$

Solving:

$9x > -18 \Rightarrow x > -2$

For the turning point analysis, we identify where:

$f'(x) = 3x^2 - 8x - 3 < 0$

The critical points allow us to test intervals to find for which x-values f'(x) is negative.