A function is not always given by an explicit formula. Instead, a problem may give a relationship such as $f(3x-1)=x^2+2x$.

This tells us the value of $f$ at the input $3x-1$. We can use the relationship without first finding a formula for $f(x)$.

However, not every relationship determines the function at every input. We need to establish which function values are determined by the information, whether different values of $x$ give consistent results, and whether the required value can actually be found.

1. Finding a Value from an Indirect Relationship

Suppose $f(3x-1)=x^2+2x$ for all real $x$, and we want to find $f(8)$.

The input to $f$ is $3x-1$. To obtain $f(8)$, set $3x-1=8$.

This gives $x=3$, so $f(8)=3^2+2(3)=15$.

There is no need to rearrange the original relationship into a formula for $f(x)$.

More generally, if $f(A(x))=B(x)$, then to find $f(c)$, we first solve $A(x)=c$.

If a solution exists, the corresponding value of $B(x)$ gives the required function value, provided the relationship is consistent.

The equation $A(x)=c$ may have one solution, several solutions or no solution. These cases require different interpretations.

2. Which Function Values Are Determined?

Consider a function $f$ defined for all real numbers and satisfying $f(x^2+1)=x^2+3$ for every real $x$.

To find $f(5)$, we set $x^2+1=5$.

This gives $x=\pm2$. Both values give $f(5)=4+3=7$.

But what if we want to find $f(0)$?

We would need $x^2+1=0$, which has no real solution.

The original relationship never gives us information about $f$ at an input of $0$.

This does not mean that $f(0)$ is undefined. The function is defined for all real numbers, but the given information does not determine its value at $0$.

The range of the given input

The expression $x^2+1$ can take every real value greater than or equal to $1$, but no value below $1$.

Let $t=x^2+1$.

Then $t\geq1$ and $x^2=t-1$. The original relationship becomes

$$\begin{aligned} f(t)&=x^2+3\\ &=t+2. \end{aligned}$$

Therefore, $f(t)=t+2\qquad\text{for }t\geq1$.

Notice that this result applies only when $t\geq1$.

We cannot simply extend the formula $f(t)=t+2$ to negative inputs.

For example, two functions could both satisfy the original relationship while having $f(0)=0$ and $f(0)=20$, respectively. They would agree for every input $t\geq1$ but differ at $t=0$.

Consequently, the given relationship does not determine a unique function on the whole real line.

When a relationship is given in the form $f(A(x))=B(x)$, first establish which values the expression $A(x)$ can take. The relationship provides information about $f$ only at those inputs.

3. Different Values Giving the Same Function Input

Sometimes the equation $A(x)=c$ has several solutions.

If two values of $x$ produce the same input to a function, they must also produce the same output.

Suppose $f(x^2)=x^4+2x^2$ for all real $x$.

The values $x=2$ and $x=-2$ both produce the input $4$.

Their corresponding outputs are also equal: $2^4+2(2^2)=24$ and $(-2)^4+2(-2)^2=24$.

Thus both values correctly give $f(4)=24$.

Now consider a different proposed relationship: $f(x^2)=x+4$ for all real $x$.

Taking $x=2$ gives $f(4)=6$, whereas taking $x=-2$ gives $f(4)=2$.

No function can satisfy both equations.

The problem is not that two different values of $x$ produce the same input to $f$. This is perfectly possible. The contradiction arises because the same input to $f$ is assigned two different outputs.

Using this condition to determine a parameter

Suppose a function $f$ satisfies $f((x-1)^2)=x^2+ax+1$ for every real $x$, where $a$ is a constant.

Find the value of $a$ for which such a function can exist.

The expression $(x-1)^2$ takes the same value when $x$ is replaced by $2-x$: $(x-1)^2=((2-x)-1)^2$.

Therefore, the corresponding outputs must be equal: $x^2+ax+1=(2-x)^2+a(2-x)+1$.

Expanding and simplifying gives $2(a+2)(x-1)=0$.

This must hold for every real $x$. Since $x-1$ is not always zero, $a=-2$.

But we have only established a necessary condition. We should also check that a function actually exists when $a=-2$.

Substituting this value gives

$$\begin{aligned} f((x-1)^2) &=x^2-2x+1\\ &=(x-1)^2. \end{aligned}$$

Hence the relationship requires $f(t)=t\qquad(t\geq0)$.

For example, the function $f(t)=t$, defined for all real $t$, satisfies the relationship.

Therefore $a=-2$ is both necessary and sufficient for the existence of such a function.

The original condition does not specify how $f$ must behave at negative inputs, but this does not prevent a function from existing.

A general condition

Suppose $f(A(x))=B(x)$ holds for every $x$ in a specified domain.

If two allowed values $u$ and $v$ satisfy $A(u)=A(v)$, then they must also satisfy $B(u)=B(v)$.

If this fails for even one pair, no function can satisfy the given relationship.

Conversely, if this condition holds for every such pair, then the relationship consistently defines $f$ on the range of $A$. Values outside that range can be chosen separately, provided no additional conditions are violated.

This gives a way to determine unknown constants or establish whether a proposed function relationship is possible without finding an explicit formula for the entire function.

4. What Can We Actually Conclude?

A function relationship may lead to three different conclusions:

These conclusions must not be confused.

Consider $f(x^2)=x^4+1$ for every real $x$.

The value $f(9)$ is determined. Setting $x^2=9$ gives $x=\pm3$, and both values produce $f(9)=82$.

However, $f(-1)$ is not determined because no real value of $x$ satisfies $x^2=-1$.

This is a lack of information, not a contradiction.

Now consider a different relationship: $f(x^2)=x^3+1$ for every real $x$.

Taking $x=1$ gives $f(1)=2$, while taking $x=-1$ gives $f(1)=0$.

These requirements contradict each other, so no function can satisfy this relationship for every real $x$.

Failure to obtain a required input does not mean that the function cannot exist. A contradiction occurs only when the given conditions require incompatible outputs for the same function input.

---

Worked Example — Determining a Function from a Restricted Relationship

A function $f$ is defined for all real numbers and satisfies $f(x^2-4x+5)=3x^2+px+q$ for every real $x$, where $p$ and $q$ are constants.

It is also given that $f(1)=7$.

Find:

(a) the values of $p$ and $q;

(b) the value of $f(10)$;

(c) whether the given information determines $f(0)$.

Solution

(a) Determine the constants

First, rewrite the input to $f$: $x^2-4x+5=(x-2)^2+1$.

The values $x$ and $4-x$ produce the same input because $(x-2)^2=((4-x)-2)^2$.

Their corresponding outputs must therefore be equal: $3x^2+px+q = 3(4-x)^2+p(4-x)+q$.

Expanding and simplifying, $(2p+24)(x-2)=0$.

Since this holds for every real $x$, $p=-12$.

The relationship is now $f(x^2-4x+5)=3x^2-12x+q$.

To use $f(1)=7$, find the value of $x$ that makes the input equal to $1$: $x^2-4x+5=1$.

Hence $(x-2)^2=0$, so $x=2$.

Substituting,

$$\begin{aligned} 7&=3(2)^2-12(2)+q\\ &=q-12. \end{aligned}$$

Therefore, $p=-12,\qquad q=19$.

(b) Find $f(10)$

The given relationship becomes $f(x^2-4x+5)=3x^2-12x+19$.

Completing the square on both sides gives $f((x-2)^2+1)=3(x-2)^2+7$.

Let $t=(x-2)^2+1$.

Then

$$(x-2)^2=t-1,$$

and so

$$\begin{aligned} f(t)&=3(t-1)+7\\ &=3t+4. \end{aligned}$$

Thus $f(t)=3t+4\qquad(t\geq1)$.

Since $10\geq1$, the formula applies: $f(10)=34$.

(c) Is $f(0)$ determined?

The expression $(x-2)^2+1$ has minimum value $1$. Its range is therefore $[1,\infty)$.

The given relationship determines $f(t)$ only for $t\geq1$.

It gives no information about $f(0)$.

For instance, both of the following functions satisfy all the conditions:

$$f_1(t)=3t+4$$

for all real $t$, and

$$f_2(t)= \begin{cases} 3t+4,&t\geq1,\\ 0,&t<1. \end{cases}$$

However, $f_1(0)=4,\qquad f_2(0)=0$.

Since two functions satisfying every given condition can have different values at $0$, the value of $f(0)$ is not determined.

---

When working with an indirect function relationship, identify the inputs covered by the given expression and use only the information that follows from it. If several values produce the same function input, check that their outputs agree. Do not extend a formula beyond the inputs for which it has been established.