Differentiation: Question 3

Worked solution

Part (a): Finding $\dfrac{dy}{dx}$

Differentiating term by term:

$\frac{dy}{dx} = 3x^2 - 12x + 9$

Recomputing independently, term by term: $\dfrac{d}{dx}(x^3)=3x^2$, $\dfrac{d}{dx}(-6x^2)=-12x$, $\dfrac{d}{dx}(9x)=9$, $\dfrac{d}{dx}(2)=0$. Adding these confirms

$\frac{dy}{dx} = 3x^2 - 12x + 9.$

Part (b): Finding the stationary points

Stationary points occur where $\dfrac{dy}{dx} = 0$:

$3x^2 - 12x + 9 = 0$

Divide through by $3$:

$x^2 - 4x + 3 = 0$

Factorise:

$(x - 1)(x - 3) = 0 \implies x = 1 \text{ or } x = 3$

Check the factorisation by expanding: $(x-1)(x-3) = x^2 - 3x - x + 3 = x^2 - 4x + 3$ ✓, matching the equation above.

Substitute each $x$-value back into $y = x^3 - 6x^2 + 9x + 2$:

At $x = 1$: $y = 1 - 6 + 9 + 2 = 6$, giving the point $(1, 6)$.

At $x = 3$: $y = 27 - 54 + 27 + 2 = 2$, giving the point $(3, 2)$.

So the stationary points are $(1, 6)$ and $(3, 2)$.

Part (c): Classifying each stationary point

Differentiate again:

$\frac{d^2y}{dx^2} = 6x - 12$

At $x = 1$: $\dfrac{d^2y}{dx^2} = 6(1) - 12 = -6 < 0$, so $(1, 6)$ is a maximum point.

At $x = 3$: $\dfrac{d^2y}{dx^2} = 6(3) - 12 = 6 > 0$, so $(3, 2)$ is a minimum point.

Part (d): Where the curve is decreasing

The curve is decreasing where $\dfrac{dy}{dx} < 0$, i.e. where $3(x-1)(x-3) < 0$. Since $3 > 0$, this is negative exactly between the roots:

$1 < x < 3$

Independent check using test values. Pick a value in each region and substitute into $\dfrac{dy}{dx}=3x^2-12x+9$:

• $x=0$ (left of $1$): $3(0)-0+9 = 9 > 0$ → increasing.
• $x=2$ (between $1$ and $3$): $3(4)-24+9 = 12-24+9 = -3 < 0$ → decreasing.
• $x=4$ (right of $3$): $3(16)-48+9 = 48-48+9 = 9 > 0$ → increasing.

This confirms the curve increases, then decreases between $x=1$ and $x=3$, then increases again, consistent with $(1,6)$ being a maximum and $(3,2)$ being a minimum, and with the decreasing interval $1 < x < 3$.

Final answers

• (a) $\dfrac{dy}{dx} = \boxed{3x^2 - 12x + 9}$
• (b) Stationary points: $\boxed{(1, 6)}$ and $\boxed{(3, 2)}$
• (c) $(1,6)$ is a maximum; $(3,2)$ is a minimum
• (d) Decreasing for $\boxed{1 < x < 3}$