Where a Function Is Increasing or Decreasing

“Increasing” and “decreasing” are questions about the sign of the gradient, so they reduce to solving an inequality in $\frac{dy}{dx}$. For a cubic, $\frac{dy}{dx}$ is a quadratic, so the finish is a quadratic inequality, with the stationary points as the boundaries.

Find the range of values of $x$ for which the curve $y = x^3 - 3x^2 - 9x + 5$ is increasing. [5]

The working

Step 1, differentiate: $\frac{dy}{dx} = 3x^2 - 6x - 9 \quad \text{(M1)}$

Step 2, “increasing” means $\frac{dy}{dx} > 0$: $3x^2 - 6x - 9 > 0 \;\Rightarrow\; x^2 - 2x - 3 > 0 \quad \text{(M1)}$

Step 3, factorise to find the boundaries: $(x - 3)(x + 1) > 0 \;\Rightarrow\; \text{roots at } x = 3 \text{ and } x = -1 \quad \text{(A1)}$

Step 4, read the inequality. The quadratic opens upward, so it’s positive outside the roots: $x < -1 \quad \text{or} \quad x > 3 \quad \text{(M1, A1)}$

The curve is increasing for $x < -1$ and for $x > 3$, and decreasing in between.

Where the marks are won and lost

• “Increasing” is $\frac{dy}{dx} > 0$, “decreasing” is $\frac{dy}{dx} < 0$. Reading the wrong sign inverts the answer.
• The solution is two regions joined by “or”, the outside of an upward parabola. Giving $-1 < x < 3$ (the inside) is the classic slip, and it’s actually where the curve decreases.
• Dividing by $3$ to simplify $x^2 - 2x - 3 > 0$ makes the factorising clean.

Common mistakes

• Solving $\frac{dy}{dx} = 0$ and stopping (that gives the boundaries, not the intervals).
• Giving the “between the roots” region for “increasing”.
• Sign slips in the derivative $3x^2 - 6x - 9$.

Full method: Stationary Points notes. Topic home: Calculus pillar.