Differentiation: Question 8

Worked solution

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

Differentiating term by term:

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

Recomputing independently, term by term: $\dfrac{d}{dx}(2x^3)=6x^2$, $\dfrac{d}{dx}(-3x^2)=-6x$, $\dfrac{d}{dx}(-12x)=-12$, $\dfrac{d}{dx}(5)=0$. Adding these confirms

$\frac{dy}{dx} = 6x^2 - 6x - 12.$

Part (b): Finding the stationary points

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

$6x^2 - 6x - 12 = 0$

Divide through by $6$:

$x^2 - x - 2 = 0$

Factorise:

$(x - 2)(x + 1) = 0 \implies x = 2 \text{ or } x = -1$

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

Substitute each $x$-value back into $y = 2x^3 - 3x^2 - 12x + 5$:

At $x = -1$: $y = 2(-1) - 3(1) - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12$, giving the point $(-1, 12)$.

At $x = 2$: $y = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15$, giving the point $(2, -15)$.

So the stationary points are $(-1, 12)$ and $(2, -15)$.

Part (c): Classifying each stationary point

Differentiate again:

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

At $x = -1$: $\dfrac{d^2y}{dx^2} = 12(-1) - 6 = -18 < 0$, so $(-1, 12)$ is a maximum point.

At $x = 2$: $\dfrac{d^2y}{dx^2} = 12(2) - 6 = 18 > 0$, so $(2, -15)$ is a minimum point.

Part (d): Where the curve is increasing

The curve is increasing where $\dfrac{dy}{dx} > 0$, i.e. where $6(x-2)(x+1) > 0$. Since $6 > 0$, this is positive exactly outside the roots:

$x < -1 \text{ or } x > 2$

Independent check using test values. Pick a value in each region and substitute into $\dfrac{dy}{dx}=6x^2-6x-12$:

• $x=-2$ (left of $-1$): $6(4) - 6(-2) - 12 = 24 + 12 - 12 = 24 > 0$ → increasing.
• $x=0$ (between $-1$ and $2$): $6(0) - 0 - 12 = -12 < 0$ → decreasing.
• $x=3$ (right of $2$): $6(9) - 18 - 12 = 54 - 18 - 12 = 24 > 0$ → increasing.

This confirms the curve increases, then decreases between $x=-1$ and $x=2$, then increases again, consistent with $(-1,12)$ being a maximum and $(2,-15)$ being a minimum, and with the increasing set $x<-1$ or $x>2$.

Final answers

• (a) $\dfrac{dy}{dx} = \boxed{6x^2 - 6x - 12}$
• (b) Stationary points: $\boxed{(-1, 12)}$ and $\boxed{(2, -15)}$
• (c) $(-1,12)$ is a maximum; $(2,-15)$ is a minimum
• (d) Increasing for $\boxed{x < -1 \text{ or } x > 2}$