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

From the function f(x), we start by plugging in known values:

1. Set f(-2) = 0:

$f(-2) = (-2)^3 + b(-2)^2 + c(-2) + d = 0$

This simplifies to:

$-8 + 4b - 2c + d = 0 \\ (1)$

2. Set f(0) = 18:

$f(0) = d = 18 \\ (2)$

3. Set f(3) = 0:

$f(3) = (3)^3 + b(3)^2 + c(3) + d = 0$

This simplifies to:

$27 + 9b + 3c + 18 = 0 \\ 9b + 3c + 45 = 0 \\ (3)$

Using equations (1) and (2):

Substituting d = 18 into equation (1):

$-8 + 4b - 2c + 18 = 0 \Rightarrow 4b - 2c + 10 = 0 \Rightarrow 2b - c + 5 = 0 \Rightarrow c = 2b + 5 \\ (4)$

Now using equation (3):

$9b + 3(2b + 5) + 45 = 0 \Rightarrow 9b + 6b + 15 + 45 = 0 \Rightarrow 15b + 60 = 0 \Rightarrow b = -4$

Using b = -4 in equation (4):

$c = 2(-4) + 5 = -8 + 5 = -3$

Thus, we have shown that b = -4, c = -3, and d = 18.

R is one of the turning points of f, which can be found by taking the first derivative:

$f'(x) = 3x^2 + 2bx + c$

Substituting b and c:

$f'(x) = 3x^2 + 2(-4)x + (-3) = 3x^2 - 8x - 3$

Setting the derivative equal to zero to find turning points:

$3x^2 - 8x - 3 = 0$

Using the quadratic formula:

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} = \frac{8 \pm \sqrt{64 + 36}}{6} = \frac{8 \pm 10}{6}$

This gives:

$x = \frac{18}{6} = 3 \quad \text{and} \quad x = \frac{-2}{6} = -\frac{1}{3}$

Now substituting x = 3 back into the original function f(x):

$f(3) = (3)^3 + (-4)(3)^2 + (-3)(3) + 18 = 27 - 36 - 9 + 18 = 0$

The coordinates of R are therefore (3, 0).

The slope of the tangent at R can be found using the derivative:

$f'(3) = 3(3)^2 + 2(-4)(3) - 3 = 27 - 24 - 3 = 0$

Thus, the equation of the tangent line at point R (3, 0) is a horizontal line:

$y = 0$.

For g(x) to be greater than zero:

$g(x) = 9x + 18 > 0 \Rightarrow 9x > -18 \Rightarrow x > -2$.

Thus, the values of x for which g(x) > 0 are:

$x ext{ in } (-2, \infty)$.

To find where f'(x) is less than zero, solve:

$3x^2 - 8x - 3 < 0$

Using the roots previously derived, the sign of the quadratic between the roots (approximately -0.33 and 3) is negative:

Thus:

$x ext{ in } (-0.33, 3)$.