Extremum problems on a closed and an open interval

On a closed interval $\displaystyle a\leq x\leq b$, a continuous function takes a greatest value and a least value, and each of them is at one of two kinds of places:

• at a point inside the interval where $\displaystyle {f'(x)}={0}$, or where the derivative is not defined;
• at one of the ends of the interval, $\displaystyle {x}={a}$ or $\displaystyle {x}={b}$ – an extremum at an end.

So you work out the values at all these points and compare them; no table of signs is needed.

An open interval has no ends to check, and the function may have no greatest or no least value there: it can approach a value without reaching it. There you decide the kind of each point by a table of signs of $\displaystyle f'$, and look at what the function does near the ends of the interval.

The example below finds the greatest and the least value of $\displaystyle {f(x)}={\frac{x}{x^2+4}}$, first on the interval $\displaystyle 0\leq x\leq5$ and then on the interval $\displaystyle x>0$.

• $\displaystyle {f'(x)}={\frac{1\cdot(x^2+4)-x\cdot2x}{(x^2+4)^2}}={\frac{4-x^2}{(x^2+4)^2}}$.
• $\displaystyle {f'(x)}={0}$ where $\displaystyle {x}={2}$ or $\displaystyle {x}={-2}$; only $\displaystyle {x}={2}$ is in the interval.
• The values: $\displaystyle {f(2)}={\frac{2}{8}}={\frac{1}{4}}$, $\displaystyle {f(0)}={0}$, $\displaystyle {f(5)}={\frac{5}{29}}$.

On the closed interval $\displaystyle 0\leq x\leq5$:

• the greatest value is $\displaystyle \frac{1}{4}$, at $\displaystyle {x}={2}$;
• the least value is 0, at $\displaystyle {x}={0}$ – an extremum at an end.

On the open interval $\displaystyle x>0$:

• the greatest value is still $\displaystyle \frac{1}{4}$: the function rises up to $\displaystyle {x}={2}$ and falls after it;
• there is no least value: the function is positive, and as $\displaystyle x$ approaches 0 or grows without end it approaches 0 – but never reaches it.

Two tent camps are on sandy ground, on one side of a straight road.
Camp $\displaystyle A$ is 2 km from the road, and camp $\displaystyle B$ is 1 km from the road.
$\displaystyle C$ is the point of the road nearest to $\displaystyle A$, and $\displaystyle D$ the point nearest to $\displaystyle B$; $\displaystyle {CD}={1.5}$ km.
The leaders’ car stops at a point $\displaystyle P$ between $\displaystyle C$ and $\displaystyle D$.
The pupils run from $\displaystyle A$ to the car, and from there to $\displaystyle B$ – in straight lines, on the sand only.
Where should the car stop to make the run as short as possible? And where to make it as long as possible?

The solution below writes $\displaystyle {CP}={x}$, expresses the length of the run by $\displaystyle x$, and compares the point where the derivative is 0 with the ends of the interval.

• The interval is closed: $\displaystyle 0\leq x\leq1.5$.
• By Pythagoras’ theorem: $\displaystyle {AP}={\sqrt{x^2+4}}$, $\displaystyle {PB}={\sqrt{(1.5-x)^2+1}}$, so $\displaystyle {L(x)}={\sqrt{x^2+4}+\sqrt{(1.5-x)^2+1}}$.
• $\displaystyle {L'(x)}={\frac{x}{\sqrt{x^2+4}}-\frac{1.5-x}{\sqrt{(1.5-x)^2+1}}}$.
• $\displaystyle {L'(x)}={0}$: $\displaystyle {\frac{x}{\sqrt{x^2+4}}}={\frac{1.5-x}{\sqrt{(1.5-x)^2+1}}}$. In the interval both sides are not negative, so squaring is allowed: $\displaystyle {x^2\left[(1.5-x)^2+1\right]}={(1.5-x)^2(x^2+4)}$.
• After expanding and cancelling, $\displaystyle {x^2}={4(1.5-x)^2}$ is left, that is $\displaystyle {x}={2(1.5-x)}$, and $\displaystyle {x}={1}$. (The other possibility, $\displaystyle {x}={-2(1.5-x)}$, gives $\displaystyle {x}={3}$, outside the interval.)
• The values: $\displaystyle {L(1)}={\sqrt{5}+\sqrt{1.25}\approx3.354}$; $\displaystyle {L(0)}={2+\sqrt{3.25}\approx3.803}$; $\displaystyle {L(1.5)}={\sqrt{6.25}+1}={3.5}$.

• The shortest run: the car stops 1 km from $\displaystyle C$, and the run is about 3.35 km.
• The longest run: the car stops at $\displaystyle C$ itself, and the run is about 3.80 km – an extremum at an end of the interval. At $\displaystyle {x}={1}$ the derivative is 0, but that is a minimum point; the maximum is found only by comparing with the ends.

The function $\displaystyle {f(x)}={\sqrt{x}-\frac{x}{4}}$ is defined on the interval $\displaystyle 0\leq x\leq9$.

Find $\displaystyle f'(x)$.

Solve the equation $\displaystyle {f'(x)}={0}$.

Work out the values of the function at the point you found and at both ends of the interval – a card for each value.

What is the greatest value of $\displaystyle f$ on the interval, and what is the least?
Which of them is an extremum at an end?

The function $\displaystyle {f(x)}={\frac{x^2+3}{x-1}}$ is defined on the open interval $\displaystyle x>1$.

Find $\displaystyle f'(x)$.

Solve the equation $\displaystyle {f'(x)}={0}$.

Which solution of the equation is in the interval?
Make a table of signs of $\displaystyle f'$ in the interval, and find the least value of $\displaystyle f$.
Does $\displaystyle f$ have a greatest value in the interval? Explain by what the function does near $\displaystyle {x}={1}$.

$\displaystyle P$ is a point on the graph of $\displaystyle {y}={\frac{4}{x}}$, with $\displaystyle x>0$.
You are looking for the point $\displaystyle P$ nearest to the origin.

Show that the distance from $\displaystyle P$ to the origin is $\displaystyle {d(x)}={\sqrt{x^2+\frac{16}{x^2}}}$.
Explain why it is enough to find the minimum of the expression under the root, $\displaystyle x^2+\frac{16}{x^2}$, and find its derivative.

Solve the equation that makes the derivative 0 (it has been divided by 2).

Work out the shortest distance.

Write the point $\displaystyle P$ nearest to the origin.
Which solution of the equation did you reject, and why?

Two villages, $\displaystyle A$ and $\displaystyle B$, are on one side of a straight road.
$\displaystyle A$ is 3 km from the road, and $\displaystyle B$ is 1 km from it.
$\displaystyle C$ and $\displaystyle D$ are the points of the road nearest to $\displaystyle A$ and to $\displaystyle B$; $\displaystyle {CD}={4}$ km.
A pumping station is built at a point $\displaystyle P$ of the road, between $\displaystyle C$ and $\displaystyle D$, with a straight pipe from it to each village.
$\displaystyle x$ – the distance $\displaystyle CP$, in kilometres.

Show that the length of the two pipes together is $\displaystyle {L(x)}={\sqrt{x^2+9}+\sqrt{(4-x)^2+1}}$.
What is the interval of $\displaystyle x$?

$\displaystyle {L'(x)}={\frac{x}{\sqrt{x^2+9}}-\frac{4-x}{\sqrt{(4-x)^2+1}}}$.
Solve the equation $\displaystyle {L'(x)}={0}$.

Work out $\displaystyle L$ at the point you found and at both ends of the interval.

Where is it best to build the station, and what is the least length of the pipes?
Where is the length greatest?