Maximum and Minimum with Restricted Inputs

Summary: A function may have an obvious maximum or minimum, but a TMUA problem may restrict which inputs are allowed. The maximum or minimum must be found using only those allowed inputs.

Consider $f(x)=(x-3)^2+2$.

Over all real values of $x$, the graph has its minimum at $(3,2)$.

Now compare two different restrictions on $x$.

The formula has not changed. Only the allowed inputs have changed.

So the first question should be:

Which values of $x$ am I actually allowed to use?

Find the Allowed Inputs First

A restriction may be given directly, or it may need to be extracted from another condition.

Do this before searching for a maximum or minimum.

A turning point outside the allowed interval cannot give the required maximum or minimum.

First find the allowed inputs. Then optimise only over those inputs.

The restriction does not always apply directly to $x$.

For example, if $u=5^x$, then $u>0$.

So if an expression is rewritten in terms of $u$, only positive values of $u$ can be used.

Similarly, if $-2\le f(x)\le2$ and we define $t=f(x)$, then $-2\le t\le2$.

Any maximum or minimum involving $t$ must therefore be found using only this range of values.

The same idea applies whenever a new variable is introduced: first find which values that variable can actually take.

Find the Candidates

Once the allowed interval is known, there are usually only a few points that can give the maximum or minimum.

A turning point outside the allowed interval is not a candidate.

Sometimes the behaviour of the function makes the comparison even quicker.

For a differentiable function on a closed interval, a useful approach is:

If the graph has a sharp turning point, such as the vertex of a modulus graph, include that point as well.

Stationary Points Are Only Candidates

Consider $f(x)=x^3-3x$ for $-2\le x\le2$.

Differentiating gives $f'(x)=3x^2-3$.

So the stationary points satisfy $3x^2-3=0$, giving $x=-1 \qquad\text{and}\qquad x=1$.

Both lie inside the allowed interval.

Together with the endpoints, there are four candidates.

Therefore $\text{maximum}=2$ and $\text{minimum}=-2$.

Finding the stationary points is not enough. Compare every allowed candidate.

Check Whether the Boundary Is Included

A maximum or minimum must actually be reached.

Compare the following two domains for $f(x)=x$.

For $0<x\le3$, the function can take values as close to $0$ as we like, but it can never equal $0$.

So $0$ is a lower bound, but it is not a minimum.

A maximum or minimum must actually be reached.

This distinction is important whenever a problem contains strict inequalities or excludes a boundary point.

When a Parameter Changes the Candidates

A parameter can change which points are available or which point gives the maximum or minimum.

There are two common ways this happens:

• the turning point moves;
• the allowed interval moves.

The important question is always:

As the parameter changes, which candidates are available, and which one gives the required extreme?

A Moving Turning Point

Consider $f(x)=(x-p)^2$ for $0\le x\le6$.

The vertex is at $x=p$.

So $p=0 \qquad\text{and}\qquad p=6$ are important values because they mark when the vertex enters or leaves the allowed interval.

The maximum requires a different comparison.

When $0\le p\le6$, the maximum occurs at whichever endpoint is farther from $p$.

At the endpoints, $f(0)=p^2$ and $f(6)=(6-p)^2$.

These are equal when $p^2=(6-p)^2$, which gives $p=3$.

Therefore:

Different parameter values matter for different reasons:

• $p=0$ and $p=6$ determine whether the vertex is inside the allowed interval;
• $p=3$ determines which endpoint gives the maximum.

A Moving Allowed Interval

Now consider $f(x)=(x-1)^2$ for $p\le x\le p+2$.

The function has its unrestricted minimum at $x=1$.

But the allowed interval now moves as $p$ changes.

If $p+2<1$, the whole interval lies to the left of $1$.

The allowed point closest to $1$ is the right endpoint $x=p+2$, so $m(p)=(p+1)^2$.

If $p\le1\le p+2$, then $x=1$ is allowed, so $m(p)=0$.

This happens when $-1\le p\le1$.

If $p>1$, the whole interval lies to the right of $1$.

The allowed point closest to $1$ is the left endpoint $x=p$, so $m(p)=(p-1)^2$.

Therefore

$$m(p)= \begin{cases} (p+1)^2, & p<-1,\\[4pt] 0, & -1\le p\le1,\\[4pt] (p-1)^2, & p>1. \end{cases}$$

The graph of $m(p)$ shows clearly what happens as the allowed interval moves.

For $-1\le p\le1$, the moving interval contains $x=1$.

So the minimum remains $m(p)=0$, which produces the flat part of the graph.

When $p<-1$ or $p>1$, the interval no longer contains $x=1$.

The minimum is then determined by the endpoint closest to $1$, and the minimum value increases as the interval moves farther away.

So the graph makes the change in candidates visible:

$$\boxed{ \text{turning point is allowed} \Rightarrow m(p)=0 }$$

but

$$\boxed{ \text{turning point is not allowed} \Rightarrow \text{nearest endpoint determines }m(p) }$$

The function itself has not changed. The allowed interval has moved, and this creates a new function $m(p)$.

More generally, in parameter questions look for values where:

• a turning point enters or leaves the allowed interval;
• an endpoint passes a turning point;
• two candidate values become equal;
• a different candidate begins to give the maximum or minimum.

These values often mark where the formula for the extreme value changes.

Optimising a Maximum or Minimum

Sometimes finding the maximum or minimum is only the first stage of the problem.

For each value of a parameter, we may first obtain a maximum or minimum.

That extreme value is then itself a function of the parameter.

For example, we might call it $M(p)$ for a maximum, or $m(p)$ for a minimum.

The problem may then ask how large or small this new quantity can be.

The important idea is:

$$\boxed{ \text{fix the parameter} \longrightarrow \text{find the required extreme} \longrightarrow \text{obtain a new function of the parameter} \longrightarrow \text{optimise again} }$$

This same method works whether the question asks for:

• the smallest possible maximum;
• the largest possible maximum;
• the smallest possible minimum;
• the largest possible minimum.

There is no need to learn four separate methods.

First Find the Extreme for Each Parameter Value

Return to $f(x)=(x-p)^2, \qquad 0\le x\le6$, with $0\le p\le6$.

For a fixed value of $p$, the maximum occurs at whichever endpoint is farther from $p$.

At the endpoints, $f(0)=p^2$ and $f(6)=(6-p)^2$.

For $0\le p\le3$, the farther endpoint is $x=6$, so $M(p)=(6-p)^2$.

For $3\le p\le6$, the farther endpoint is $x=0$, so $M(p)=p^2$.

Therefore

$$M(p)= \begin{cases} (6-p)^2, & 0\le p\le3,\\[4pt] p^2, & 3\le p\le6. \end{cases}$$

The first stage is complete.

Seeing the Extreme Value as a New Graph

The original function was a function of $x$.

The new function $M(p)$ records the maximum value obtained for each value of $p$.

For every value of $p$, the height of this graph is the maximum value of the original function.

Now we can optimise the new graph.

Its lowest point occurs at $p=3$, where $M(3)=9$.

Therefore the smallest possible maximum is $9$.

The largest values occur at the ends of the parameter range: $M(0)=M(6)=36$.

Therefore the largest possible maximum is $36$.

So the two stages can be viewed as

$$\boxed{ f(x) \longrightarrow M(p)\text{ or }m(p) \longrightarrow \text{optimise the new function} }$$

The same idea applies if the first stage gives a minimum rather than a maximum.

Do not try to optimise everything at once. First find the required maximum or minimum for each parameter value. Then optimise those extreme values.

The second stage also needs the usual boundary check.

If the allowed range of the parameter has an excluded endpoint, the new function $M(p)$ or $m(p)$ may approach a value without ever reaching it.

What Should You Look For?

When the first-stage extreme depends on a parameter, important parameter values often occur when:

• the point giving the extreme changes;
• two competing candidate values become equal;
• a stationary or turning point enters or leaves the allowed region;
• the parameter reaches an endpoint of its own allowed range.

These are the natural places to check when optimising the extreme value itself.

When an Extreme Value Becomes a Condition

Sometimes the question does not ask you to optimise $M(p)$ or $m(p)$.

Instead, it tells you that the extreme value must satisfy a condition.

Return again to $f(x)=(x-p)^2, \qquad 0\le x\le6$, where $0\le p\le6$.

We already know that

$$M(p)= \begin{cases} (6-p)^2, & 0\le p\le3,\\[4pt] p^2, & 3\le p\le6. \end{cases}$$

Suppose the maximum is required to satisfy $M(p)\le16$.

The graph gives a useful picture of the condition.

We want the part of the graph at or below the level $M=16$.

The graph meets this level at $p=2 \qquad\text{and}\qquad p=4$.

So the graph suggests $2\le p\le4$.

Now confirm this algebraically.

For $0\le p\le3$, we need $(6-p)^2\le16$.

Since $6-p$ is positive in this range, $6-p\le4$, so $p\ge2$.

Together with the case condition, $2\le p\le3$.

For $3\le p\le6$, we need $p^2\le16$, so $p\le4$.

Together with the case condition, $3\le p\le4$.

Combining the two cases gives $2\le p\le4.$

The structure is

$$\boxed{ \text{find the extreme} \longrightarrow \text{express it as a function of the parameter} \longrightarrow \text{apply the condition} \longrightarrow \text{solve for the parameter} }$$

This works with conditions such as $M(p)<k, \qquad M(p)\ge k, \qquad m(p)>k, \qquad m(p)\le k$.

If the extreme value has different formulas in different cases, there is one final check:

After solving a case, keep only the parameter values that actually belong to that case.

For example, a solution obtained from the formula for $0\le p\le3$ cannot be used if it gives $p=5$.

The graph helps show the structure, while the algebra confirms the exact answer.

Worked Example

For real $x$ satisfying $x^2-4x\le0$, let $f(x)=x^2-6x+13$.

Find the difference between the maximum and minimum values of $f(x)$.

Solution

First find the allowed inputs.

Since $x^2-4x=x(x-4)$, the condition $x(x-4)\le0$ gives $0\le x\le4$.

Now rewrite the function as $f(x)=(x-3)^2+4$.

The vertex occurs at $x=3$, which lies inside the allowed interval.

The candidates are therefore:

Hence $\text{maximum}=13$ and $\text{minimum}=4$.

Therefore the required difference is $13-4=\boxed{9}$.

The key step was finding the allowed interval first. Without the restriction $0\le x\le4$, the optimisation problem would be different.

Practical Approach

For a maximum or minimum problem with restricted inputs: