5.6 Stationary pointsIB Maths: Applications and Interpretation SL: Revision notes

The derivative $f'(x)$ gives the gradient of the tangent to the curve $y=f(x)$. A stationary point occurs at any value of $x$ where $f'(x)=0$, because the tangent is horizontal there. To find them, differentiate, set $f'(x)=0$ and solve. For $f(x)=x^2-6x+5$: $f'(x)=2x-6=0$, so $x=3$ and $y=f(3)=-4$, giving the stationary point $(3,-4)$. Remember the power rule: if $f(x)=ax^n$ then $f'(x)=anx^{n-1}$, and a constant differentiates to 0.

In examinations you may use your GDC rather than differentiating by hand. Most GDCs and graphing software can plot the graph of $f'(x)$ directly, or give the gradient at any chosen point. The solutions of $f'(x)=0$ are the $x$-intercepts (zeros) of the graph of $f'$, which you can read using the GDC’s zero or solve function. Example: for $f(x)=2x^3-9x^2+12x$ the derivative is $f'(x)=6x^2-18x+12$, whose graph crosses the $x$-axis at $x=1$ and $x=2$. Then use $f$ to find the $y$-values: $f(1)=5$ and $f(2)=4$. Give answers exactly or to 3 significant figures.

A local maximum is a point where $f(x)$ is greater than at all nearby points; a local minimum is lower than all nearby points. Both are stationary points, so $f'(x)=0$. To decide which, look at the sign of $f'(x)$ just before and just after the point:

• positive then negative (the curve rises, then falls): local maximum;
• negative then positive (the curve falls, then rises): local minimum. Example: $f'(x)=(x-1)(x+3)$ has $f'(-4)=5$, $f'(-2)=-3$ and $f'(0)=-3$, $f'(2)=5$. So $x=-3$ is a local maximum and $x=1$ is a local minimum.

State a stationary point as coordinates $(x,y)$. Once you know the $x$-value, substitute into the original function $f$ to get the maximum or minimum value. For $P(x)=-x^3+9x^2-15x+10$, $P'(x)=-3(x-1)(x-5)$, so the stationary points are at $x=1$ and $x=5$. $P'(4)=9>0$ and $P'(6)=-15<0$, so $x=5$ is a local maximum and $P(5)=35$. In context, interpret the answer with units: a local maximum profit of 35 thousand USD.

A local maximum is only the highest point nearby; it is not necessarily the greatest value of the function on the given domain. Likewise a local minimum is not necessarily the least value. To find the greatest or least value on a domain $a\le x\le b$, compare the values at the stationary points with $f(a)$ and $f(b)$ at the ends. Example: $T(t)=-0.02t^3+0.6t^2-3.6t+20$ for $0\le t\le24$ has a local minimum $T(3.68)=13.9$, but $T(24)=2.72$ is smaller, so the least temperature is at the end of the domain.

A ball’s height is modelled by $h(t)=-5t^2+20t+1$ metres for $0\le t\le 5$. Find the greatest height. $h'(t)=-10t+20=0$ gives $t=2$. $h'(1)=10>0$ and $h'(3)=-10<0$, so it is a local maximum. $h(2)=21$ m. Check the ends: $h(0)=1$ and $h(5)=-24$, so 21 m is the greatest height on the domain. The negative value $h(5)$ shows the ball has hit the ground before $t=5$, which is why the domain should always be checked against the context.