Stationary points and increasing and decreasing functionsAQA A-Level Maths: Revision notes

A function is increasing where its gradient is positive, $\frac{dy}{dx}>0$, and decreasing where $\frac{dy}{dx}<0$. To find the intervals, differentiate and solve the inequality. For $y=x^3-6x^2+9x+2$, $\frac{dy}{dx}=3(x-1)(x-3)$. This is positive for $x<1$ or $x>3$ (increasing) and negative for $1<x<3$ (decreasing). Because the quadratic is positive outside its roots, you can read the intervals from a quick sketch of $\frac{dy}{dx}$.

A stationary point is a point on a curve where the gradient is zero, $\frac{dy}{dx}=0$. To find them: differentiate, solve $\frac{dy}{dx}=0$ for $x$, then substitute each $x$ into the equation of the curve to get $y$. For $y=x+\frac{16}{x}$ ($x>0$), $\frac{dy}{dx}=1-\frac{16}{x^2}=0$ gives $x=4$ and $y=8$. A stationary point may be a local maximum, a local minimum or a stationary point of inflection.

The second derivative $\frac{d^2y}{dx^2}$ is the rate of change of the gradient. Where $\frac{d^2y}{dx^2}>0$ the gradient is increasing; where it is negative the gradient is decreasing. At a stationary point:

• $\frac{d^2y}{dx^2}>0$: minimum
• $\frac{d^2y}{dx^2}<0$: maximum
• $\frac{d^2y}{dx^2}=0$: inconclusive; the point could be a maximum, minimum or point of inflection. For $y=x^3-6x^2+9x+2$, $\frac{d^2y}{dx^2}=6x-12$. At $x=1$ it is $-6<0$ (maximum, $y=6$) and at $x=3$ it is $6>0$ (minimum, $y=2$).

If $\frac{d^2y}{dx^2}=0$ at a stationary point, test the sign of $\frac{dy}{dx}$ just either side. Gradient positive then negative: maximum. Negative then positive: minimum. The same sign on both sides: stationary point of inflection. The sign test works for every stationary point, so it is also a valid alternative to the second derivative whenever you are asked to justify the nature.

In context problems, form an expression for the quantity in one variable, differentiate, set the derivative to zero and justify the maximum or minimum. For an open box with square base $x$ cm and height $h$ cm made from $300$ cm$^2$ of card: $x^2+4xh=300$ gives $h=\frac{300-x^2}{4x}$, so $V=75x-\frac{x^3}{4}$. Then $\frac{dV}{dx}=75-\frac{3x^2}{4}=0$ gives $x=10$, and $\frac{d^2V}{dx^2}=-15<0$, so $V=500$ cm$^3$ is a maximum. Check that your $x$-value is possible (positive here) and give units.