5.10 Second derivative and concavityIB Maths: Applications and Interpretation HL: Revision notes

The second derivative is the derivative of the derivative. It has two notations: $\frac{d^2y}{dx^2}\quad\text{or}\quad f''(x).$ It measures how quickly the gradient is changing. Example: $f(x)=x^3-6x^2+9x$ gives $f'(x)=3x^2-12x+9$ and $f''(x)=6x-12$. Use the rules from 5.9 (chain, product, quotient) to find $f'$ first, simplify, then differentiate again. For $c(t)=4te^{-t/2}$: $c'(t)=(4-2t)e^{-t/2}$ and $c''(t)=(t-4)e^{-t/2}$.

The sign of $f''(x)$ describes the shape of the graph:

• $f''(x)>0$: the gradient is increasing and the graph is concave-up (it holds water, like a cup).
• $f''(x)<0$: the gradient is decreasing and the graph is concave-down (like a cap). In context, concave-up means the rate of change is increasing and concave-down means it is decreasing. A profit graph that is increasing and concave-down is still rising, but more slowly each year.

At a stationary point, where $f'(x)=0$:

• $f''(x)>0$ gives a local minimum.
• $f''(x)<0$ gives a local maximum.
• $f''(x)=0$ is inconclusive: check the sign of $f'$ either side. Example: $f(x)=x^3-6x^2+9x$. $f'(x)=3(x-1)(x-3)=0$ gives $x=1$ or $x=3$. $f''(1)=-6<0$, so a maximum with $f(1)=4$. $f''(3)=6>0$, so a minimum with $f(3)=0$. Always state the sign of $f''$ as your reason.

A point of inflexion is a point where the concavity changes, so $f''$ changes sign. Solve $f''(x)=0$, then check that $f''$ has a different sign either side. Example: $f''(x)=6x-12=0$ gives $x=2$. For $x<2$, $f''<0$; for $x>2$, $f''>0$. So $(2,\,2)$ is a point of inflexion, since $f(2)=8-24+18=2$. $f''(x)=0$ alone is not enough: $f(x)=x^4$ has $f''(0)=0$ but $f''(x)=12x^2\ge0$ on both sides, so there is no inflexion. In context, the point of inflexion on a model is where the rate of change is greatest or least, such as the time when a population grows fastest or a concentration falls fastest.

For motion in a straight line with displacement $s(t)$: velocity $v=\frac{ds}{dt}$ and acceleration $a=\frac{d^2s}{dt^2}$ (developed in kinematics, 5.13). Acceleration zero means the velocity is at a maximum or minimum. The second derivative also appears in second order differential equations (5.18). Method for context problems: find $f'(x)=0$ to locate turning points, use $f''$ to classify them, solve $f''(x)=0$ for inflexions, and interpret every result with units and the real-world meaning.