Grade 12 Mathematics
(a) Find $f'(x)$:
$$\begin{align*} f(x) &= x^3 - 3x + 2 \\ f'(x) &= 3x^2 - 3 \end{align*}$$(b) Find turning points by setting $f'(x) = 0$:
$$\begin{align*} 3x^2 - 3 &= 0 \\ 3x^2 &= 3 \\ x^2 &= 1 \\ x &= \pm 1 \end{align*}$$Therefore, the turning points occur at $x = -1$ and $x = 1$.
(c) Function is increasing when $f'(x) > 0$:
$$\begin{align*} 3x^2 - 3 &> 0 \\ 3x^2 &> 3 \\ x^2 &> 1 \\ x &< -1 \text{ or } x > 1 \end{align*}$$Therefore, $f(x)$ is increasing for $x \in (-\infty, -1) \cup (1, \infty)$ or $x < -1$ or $x > 1$.
(d) Function is decreasing when $f'(x) < 0$:
Therefore, $f(x)$ is decreasing for $x \in (-1, 1)$ or $-1 < x < 1$.
(a) Find $g'(x)$:
$$\begin{align*} g(x) &= 2x^3 - 6x^2 + 5 \\ g'(x) &= 6x^2 - 12x \end{align*}$$(b) Find turning points:
$$\begin{align*} g'(x) &= 0 \\ 6x^2 - 12x &= 0 \\ 6x(x - 2) &= 0 \\ x &= 0 \text{ or } x = 2 \end{align*}$$The turning points occur at $x = 0$ and $x = 2$.
(c) Function is increasing when $g'(x) > 0$:
$$\begin{align*} 6x(x - 2) &> 0 \end{align*}$$Testing intervals: $x < 0$ or $x > 2$
Therefore, $g(x)$ is increasing for $x \in (-\infty, 0) \cup (2, \infty)$ or $x < 0$ or $x > 2$.
(d) Function is decreasing when $g'(x) < 0$:
Therefore, $g(x)$ is decreasing for $x \in (0, 2)$ or $0 < x < 2$.
(a) Calculate $h'(x)$:
$$\begin{align*} h(x) &= -x^3 + 12x - 5 \\ h'(x) &= -3x^2 + 12 \end{align*}$$(b) Find where gradient is zero:
$$\begin{align*} h'(x) &= 0 \\ -3x^2 + 12 &= 0 \\ -3x^2 &= -12 \\ x^2 &= 4 \\ x &= \pm 2 \end{align*}$$The gradient is zero at $x = -2$ and $x = 2$.
(c) Function is increasing when $h'(x) > 0$:
$$\begin{align*} -3x^2 + 12 &> 0 \\ -3x^2 &> -12 \\ x^2 &< 4 \\ -2 &< x < 2 \end{align*}$$Therefore, $h(x)$ is increasing for $x \in (-2, 2)$ or $-2 < x < 2$.
(d) Function is decreasing when $h'(x) < 0$:
Therefore, $h(x)$ is decreasing for $x \in (-\infty, -2) \cup (2, \infty)$ or $x < -2$ or $x > 2$.
(a) Determine $f'(x)$:
$$\begin{align*} f(x) &= x^3 + 3x^2 - 9x + 1 \\ f'(x) &= 3x^2 + 6x - 9 \end{align*}$$(b) Calculate turning points:
$$\begin{align*} f'(x) &= 0 \\ 3x^2 + 6x - 9 &= 0 \\ x^2 + 2x - 3 &= 0 \\ (x + 3)(x - 1) &= 0 \\ x &= -3 \text{ or } x = 1 \end{align*}$$(c) Nature of turning points:
At $x = -3$: Maximum turning point (changes from increasing to decreasing)
At $x = 1$: Minimum turning point (changes from decreasing to increasing)
(d) $f$ is increasing when $f'(x) > 0$:
Therefore, $f(x)$ is increasing for $x \in (-\infty, -3) \cup (1, \infty)$ or $x < -3$ or $x > 1$.
(a) Find $p'(x)$:
$$\begin{align*} p(x) &= -2x^3 + 3x^2 + 12x - 7 \\ p'(x) &= -6x^2 + 6x + 12 \end{align*}$$(b) Solve $p'(x) = 0$:
$$\begin{align*} -6x^2 + 6x + 12 &= 0 \\ -6(x^2 - x - 2) &= 0 \\ x^2 - x - 2 &= 0 \\ (x - 2)(x + 1) &= 0 \\ x &= 2 \text{ or } x = -1 \end{align*}$$The critical points are at $x = -1$ and $x = 2$.
(c) Function is decreasing when $p'(x) < 0$:
$$\begin{align*} -6x^2 + 6x + 12 &< 0 \\ -6(x - 2)(x + 1) &< 0 \\ (x - 2)(x + 1) &> 0 \end{align*}$$Testing intervals: $x < -1$ or $x > 2$
Therefore, $p(x)$ is decreasing for $x \in (-\infty, -1) \cup (2, \infty)$ or $x < -1$ or $x > 2$.
(a) Calculate $k'(x)$:
$$\begin{align*} k(x) &= \frac{1}{3}x^3 - 2x^2 + 3x + 1 \\ k'(x) &= x^2 - 4x + 3 \end{align*}$$(b) Find turning points:
$$\begin{align*} k'(x) &= 0 \\ x^2 - 4x + 3 &= 0 \\ (x - 1)(x - 3) &= 0 \\ x &= 1 \text{ or } x = 3 \end{align*}$$The turning points occur at $x = 1$ and $x = 3$.
(c) Intervals of increase and decrease:
When $k'(x) > 0$: $(x - 1)(x - 3) > 0$
(a) Find $\frac{dy}{dx}$:
$$\begin{align*} y &= x^3 - 6x^2 + 9x + 2 \\ \frac{dy}{dx} &= 3x^2 - 12x + 9 \end{align*}$$(b) Where is the gradient zero?
$$\begin{align*} 3x^2 - 12x + 9 &= 0 \\ x^2 - 4x + 3 &= 0 \\ (x - 1)(x - 3) &= 0 \\ x &= 1 \text{ or } x = 3 \end{align*}$$(c) At $x = 0$:
$$\begin{align*} \frac{dy}{dx}\bigg|_{x=0} &= 3(0)^2 - 12(0) + 9 \\ &= 9 > 0 \end{align*}$$Since the derivative is positive, the function is increasing at $x = 0$.
(d) Function is increasing when $\frac{dy}{dx} > 0$:
$$\begin{align*} 3x^2 - 12x + 9 &> 0 \\ 3(x - 1)(x - 3) &> 0 \\ (x - 1)(x - 3) &> 0 \end{align*}$$Therefore, the function is increasing for $x \in (-\infty, 1) \cup (3, \infty)$ or $x < 1$ or $x > 3$.
(a) Determine $a$ and $b$:
If $f(x) = x^3 + ax^2 + bx + c$, then $f'(x) = 3x^2 + 2ax + b$
Given that $f'(x) = 3x^2 - 12x + 9$, we can compare coefficients:
$$\begin{align*} 2a &= -12 \quad \Rightarrow \quad a = -6 \\ b &= 9 \end{align*}$$(b) Find turning points when $c = 5$:
So $f(x) = x^3 - 6x^2 + 9x + 5$
$$\begin{align*} f'(x) &= 0 \\ 3x^2 - 12x + 9 &= 0 \\ (x - 1)(x - 3) &= 0 \\ x &= 1 \text{ or } x = 3 \end{align*}$$Find $y$-coordinates:
$$\begin{align*} f(1) &= (1)^3 - 6(1)^2 + 9(1) + 5 = 1 - 6 + 9 + 5 = 9 \\ f(3) &= (3)^3 - 6(3)^2 + 9(3) + 5 = 27 - 54 + 27 + 5 = 5 \end{align*}$$The turning points are at $(1, 9)$ and $(3, 5)$.
(c) $f$ is decreasing when $f'(x) < 0$:
$$\begin{align*} 3x^2 - 12x + 9 &< 0 \\ 3(x - 1)(x - 3) &< 0 \end{align*}$$Therefore, $f$ is decreasing for $x \in (1, 3)$ or $1 < x < 3$.
(a) Determine $g'(x)$:
$$\begin{align*} g(x) &= -x^3 + 6x^2 - 9x + 4 \\ g'(x) &= -3x^2 + 12x - 9 \end{align*}$$(b) Calculate stationary points:
$$\begin{align*} g'(x) &= 0 \\ -3x^2 + 12x - 9 &= 0 \\ -3(x^2 - 4x + 3) &= 0 \\ x^2 - 4x + 3 &= 0 \\ (x - 1)(x - 3) &= 0 \\ x &= 1 \text{ or } x = 3 \end{align*}$$(c) Nature of stationary points:
Test the sign of $g'(x)$ around each point:
At $x = 1$: Minimum (changes from decreasing to increasing)
At $x = 3$: Maximum (changes from increasing to decreasing)
(d) Function is increasing when $g'(x) > 0$:
Therefore, $g$ is increasing for $x \in (1, 3)$ or $1 < x < 3$.
(a) Turning points when $f'(x) = 0$:
$$\begin{align*} 3x^2 + 6x - 24 &= 0 \\ x^2 + 2x - 8 &= 0 \\ (x + 4)(x - 2) &= 0 \\ x &= -4 \text{ or } x = 2 \end{align*}$$(b) Function is increasing when $f'(x) > 0$:
$$\begin{align*} 3x^2 + 6x - 24 &> 0 \\ 3(x + 4)(x - 2) &> 0 \end{align*}$$Therefore, $f$ is increasing for $x \in (-\infty, -4) \cup (2, \infty)$ or $x < -4$ or $x > 2$.
(c) Function is decreasing when $f'(x) < 0$:
Therefore, $f$ is decreasing for $x \in (-4, 2)$ or $-4 < x < 2$.
(d) Find $f(x)$:
Since $f'(x) = 3x^2 + 6x - 24$, we integrate to find $f(x)$:
$$\begin{align*} f(x) &= x^3 + 3x^2 - 24x + c \end{align*}$$Using $f(0) = 10$:
$$\begin{align*} f(0) &= 0 + 0 - 0 + c = 10 \\ c &= 10 \end{align*}$$Therefore, $f(x) = x^3 + 3x^2 - 24x + 10$.
(a) Show that $p'(x) = 3(x - 2)(x - 4)$:
$$\begin{align*} p(x) &= x^3 - 9x^2 + 24x - 20 \\ p'(x) &= 3x^2 - 18x + 24 \\ &= 3(x^2 - 6x + 8) \\ &= 3(x - 2)(x - 4) \end{align*}$$(b) Turning points:
$$\begin{align*} p'(x) &= 0 \\ 3(x - 2)(x - 4) &= 0 \\ x &= 2 \text{ or } x = 4 \end{align*}$$(c) Function is strictly increasing when $p'(x) > 0$:
$$\begin{align*} 3(x - 2)(x - 4) &> 0 \\ (x - 2)(x - 4) &> 0 \end{align*}$$Therefore, $p$ is strictly increasing for $x \in (-\infty, 2) \cup (4, \infty)$ or $x < 2$ or $x > 4$.
(d) Explanation:
A stationary point where the function changes from increasing to increasing would be called a point of inflection with horizontal tangent. For cubic functions with two distinct turning points (like this one), the function must change from increasing to decreasing (maximum) or decreasing to increasing (minimum) at each turning point. The function cannot have a stationary point where it remains increasing on both sides because the derivative would need to touch zero without changing sign, which doesn't occur at either $x = 2$ or $x = 4$ for this function.
(a) Critical values:
$$\begin{align*} h'(x) &= 0 \\ -3x^2 + 12x - 9 &= 0 \\ -3(x^2 - 4x + 3) &= 0 \\ x^2 - 4x + 3 &= 0 \\ (x - 1)(x - 3) &= 0 \\ x &= 1 \text{ or } x = 3 \end{align*}$$(b) Sign diagram for $h'(x) = -3(x - 1)(x - 3)$:
Test values in each interval:
(c) Intervals of increase and decrease:
(d) Nature of turning points: