Evaluate nonlinear functions and recover inputs

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Difficulty: Intermediate
Estimated time: 38 minutes
Domains: Advanced Math
Techniques: Function-evaluationReverse-input-recoveryDomain-restrictionsParameter-recoveryFunction-composition

What you’ll learn

Put an input into a polynomial, exponential, radical, or rational function and find the output.
Work backward from a given output and find every input that gives it.
Throw out any input the original function doesn't allow, like one that makes its denominator zero, even after you simplify.
Find the exact numbers in a given form such as $f(x)=ab^x+c$ .
Handle a hard pattern the SAT keeps coming back to: one function placed inside another.

Prerequisites

You only need signed arithmetic, order of operations, and the fact that

u^2=k

has two real solutions when

k>0

Why this matters on the SAT

Decide whether the input or output is missing

SAT function questions usually give you one side, an input or an output, and ask for the other. So before any algebra, ask: which way is this question going? Then keep every answer that fits the original conditions.

Solution to the example

You're given the output, $18$ , so set the rule equal to $18$ :

(a-3)^2+2=18.

Subtract $2$ and you get $(a-3)^2=16$ . Two numbers square to $16$ , so $a-3=4$ or $a-3=-4$ . That gives $a=7$ or $a=-1$ . The question also says $a<3$ , which rules out $7$ . So $a=-1$ , and the answer is A.

This question runs backward through the function: you know the output and need the input. Solving gave you two possible inputs, and the condition $a<3$ picked the one that works.

Common mistake:

Keeping only $a=7$ after taking the square root. It’s an easy slip, because $4$ is the square root most people think of first. But $u^2=16$ has two real solutions, $u=4$ and $u=-4$ . Write both down before you apply any restriction. Stopping at $7$ leads straight to choice C, the trap.

SAT example

The function $f$ is defined by

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

If $f(a)=18$ and $a<3$ , what is the value of $a$ ?

A
$-1$
B
$1$
C
$7$
D
$19$

Read the roles before calculating

Function notation links an input to an output, and each symbol has its own job.

Try it yourself:

Look for function notation with a known input or a known output. If the question gives $f(-2)$ , you know the input, so go forward to the output. If it gives $f(a)=11$ , you know the output, so go backward to every input that works. A nested expression like $p(p(x))$ makes the same forward-or-backward choice, one layer at a time.

What each symbol in function notation means

Symbol	Job	Read it as
$f$	Function name	The rule you apply
$x$	Input	A stand-in for any allowed number you put into the rule
$f(x)$	Output	What comes out after $x$ goes through the rule
$f(a)=k$	Input and output together	Put in $a$ , get out $k$

The wording tells you which way to go:

Ask what the question gives you and what it wants.

Input to output: evaluate

You know $f(x)$ and an input such as $-2$ . Replace every $x$ with $(-2)$ and simplify.
The input is known. The output is what you find.

Output to input: solve

You know $f(a)=k$ . Set the function rule equal to $k$ and solve for every possible $a$ .
The output is known. One input may work, or several may.

Whatever sits inside the parentheses is the input, all of it: in $f(2x+1)$ , the input is $2x+1$ , not $x$ . At the end, keep only the solutions the original function and the question's conditions allow.

Evaluate a stated input

Evaluating is substitution: $f(3)$ means "put $3$ into the rule and see what comes out." Replace every copy of the variable with the input, put the input in parentheses if it's negative or a fraction, and then follow the usual order of operations.

Write the rule down first. Keep its parentheses, exponents, roots, fractions, and any restriction, so nothing gets lost once the numbers go in.
Put in the whole input. For $p(-2)$ , write $(-2)$ everywhere the rule has $x$ , including any $x$ in a denominator.
Simplify. Exponents come before multiplication and addition, so $2(-2)^2$ is $2\cdot4=8$ . Multiplying first would give $(-4)^2=16$ .
Check the domain. A denominator can't be zero, and an even root, like a square root, can't have a negative number under it. So $\sqrt{x+9}$ has no real value at $x=-10$ , because that puts $-1$ under the root.

It works the same way for every family of functions:

One evaluation in each function family

Family and rule	Input	Evaluation
Polynomial: $p(x)=2x^2-3x+1$	$-2$	$p(-2)=2(-2)^2-3(-2)+1=15$
Exponential: $e(x)=5\cdot2^x+1$	$3$	$e(3)=5\cdot2^3+1=41$
Radical: $r(x)=\sqrt{x+9}-2$	$7$	$r(7)=\sqrt{16}-2=2$
Rational: $s(x)=\dfrac{x^2-4}{x+3}$	$-1$	$s(-1)=\dfrac{1-4}{2}=-\dfrac32$

Why the parentheses matter. Take $p(x)=2x^2-3x+1$ from the table. With parentheses, $p(-2)$ starts as $2(-2)^2-3(-2)+1$ , which is $8+6+1=15$ . Without them, you'd write $2-2^2-3-2+1$ and get $-6$ . The $2(-2)^2$ turned into $2-2^2$ , and the $-3(-2)$ turned into $-3-2$ , so both products became subtractions.

Check your understanding:

If $q(x)=3^x-(x-1)^2$ , what is $q(-1)$ ?

For one short substitution like these, hand work is usually faster, and you see any domain problem, like a zero denominator, as you go.

Define once for repeated or nested evaluation

When you'll use the same rule more than once, Desmos saves you work. Define the rule once instead of typing the formula again:

p(x)=x^2-1
p(-2)
p(p(-2))

Desmos gives $p(-2)=3$ , then feeds that output back in as the next input:

p(p(-2))=p(3)=8.

The nested parentheses keep the inside-out order for you. Defining once helps most when the input is negative or a fraction, the rule is long, you need several outputs, or one function sits inside another. Keep the definition on its own line, where you can check it, and ask for each value on a new line below it. Every value then comes from the same rule, so you only have to type it correctly once.

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Define p once, then reuse its name for a direct evaluation and a nested one.

Find every input that works

When you're given an output, turn the function statement into an equation. In the opening, $f(a)=18$ became $(a-3)^2+2=18$ : the rule, with $a$ in place of $x$ , set equal to the output. In general:

f(a)=k \quad\Longrightarrow\quad \text{function rule with }a=k.

Watch the word every. A nonlinear function can send two or more inputs to the same output. You saw this in the opening: $a=7$ and $a=-1$ both give $f(a)=18$ .

Write the equation. Replace $f(a)$ with the actual rule, using $a$ as the input.
Get it into a shape you can solve. Factor, isolate a radical, use a common exponential base, or clear a denominator. Then stop rewriting: once you reach $(x-1)^2=9$ , take the square root instead of multiplying out the square.
Solve every branch. Even powers and factored products often split the equation into more than one case, or branch. $(x-1)^2=9$ splits into $x-1=3$ and $x-1=-3$ , and each one gives an answer.
Test each solution in the original. Throw out any input that makes a denominator zero or puts a negative number under an even root. If you squared both sides to get rid of a root, be extra careful, because squaring can add an answer that doesn't work: squaring $\sqrt{x}=-2$ gives $x=4$ , but $\sqrt4$ is $2$ , not $-2$ . Last, throw out anything that breaks a condition in the question, the way $a<3$ threw out $7$ in the opening.

Here's the move for each family:

Working backward in each function family

Family	Equation	Exact solution
Polynomial	$(x-1)^2+4=13$	$(x-1)^2=9$ , so $x=-2$ or $x=4$
Exponential	$2^{x+1}+3=35$	$2^{x+1}=32=2^5$ , so $x=4$
Radical	$\sqrt{3x+1}-2=5$	$\sqrt{3x+1}=7$ , so $x=16$ . That's allowed, since the domain is $x\ge-\dfrac13$
Rational	$\dfrac{12}{x+1}+2=5$	$12=3(x+1)$ , so $x=3$ . That's allowed, since only $x=-1$ is ruled out

Common mistake:

Found only one input? Go back to the line where you squared, factored, or took a root, and look for the branch you skipped. Found an extra one? Go back to the original denominator, the radical, and any conditions in the question. Your final answers have to pass both checks.

Check your understanding:

The function $p$ is defined by $p(x)=2-(x+1)^2$ . If $p(t)=-7$ , what is the sum of all possible values of $t$ ?

Example: use a graph to locate, then algebra to finish

Worked example

The function $r$ is defined by

r(x)=x+\frac{6}{x+1}

for $x>0$ . What is the sum of all values of $a$ for which $r(a)=5$ ?

Locate every solution

Start with Desmos, so you know how many solutions to look for. Enter the function with its restriction, and the target:

r(x)=x+6/(x+1){x>0}
y=5

Select both intersections. Their $x$ -coordinates are about $0.268$ and $3.732$ . So the equation has two positive solutions, and you know not to stop after one.

Get the exact values

The graph tells you how many solutions there are and roughly where. Algebra gives the exact values:

a+\frac{6}{a+1}=5.

Since $a>0$ , the denominator $a+1$ can't be zero, so it's safe to multiply both sides by $a+1$ :

a(a+1)+6=5(a+1).

Expand and collect everything on one side:

a^2-4a+1=0.

This doesn't factor nicely, so complete the square. Move the $1$ over, then add $4$ to both sides:

\begin{aligned} a^2-4a&=-1\\[1.4em] a^2-4a+4&=3\\[1.4em] (a-2)^2&=3. \end{aligned}

Take the square root and keep both signs, as in the opening: $a-2=\sqrt3$ or $a-2=-\sqrt3$ . So

a=2-\sqrt3 \quad\text{or}\quad a=2+\sqrt3.

Both values are positive, so both count. Their sum is

(2-\sqrt3)+(2+\sqrt3)=\boxed{4}.

Read the graph and the algebra together. The two intersections kept you from stopping after one solution. But the decimals aren't the final answer: the exact values are radicals, and the $\sqrt3$ parts cancel when you add them.

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Select both intersections. The worked solution then finds their exact values.

This is a hybrid problem: Desmos and algebra each did a job. Remember the split: the graph counts, the algebra finishes. The algebra is also where you check the domain, as when $a>0$ told you that $a+1$ couldn't be zero.

Find the parameters of a given form

Sometimes you already know the function's form, like

f(x)=ab^x+c,

where $a\ne0$ , $b>0$ , and $b\ne1$ , and your job is to find the numbers $a$ , $b$ , and $c$ . These fixed numbers are the function's parameters, and each one does a different job:

$a$ scales the exponential part.
$b$ is the factor the exponential part gets multiplied by each time $x$ goes up by $1$ .
$c$ shifts every output by the same amount.

That last point is the key. Since $c$ is added to every output, it cancels when you subtract one output from the next.

Here's an example. Suppose you know three outputs:

Three outputs of $f(x)=ab^x+c$

$x$	$f(x)$
$0$	$7$
$1$	$11$
$2$	$23$

Find the first two changes, from each output to the next:

D_0=f(1)-f(0)=11-7=4

and

D_1=f(2)-f(1)=23-11=12.

Why does this help? From the form, $f(0)=a+c$ , $f(1)=ab+c$ , and $f(2)=ab^2+c$ . Subtracting cancels the $c$ each time:

D_0=a(b-1) \quad\text{and}\quad D_1=ab(b-1).

So $D_1$ is $D_0$ times $b$ , and dividing $D_1$ by $D_0$ leaves $b$ :

b=\frac{D_1}{D_0}=\frac{12}{4}=3.

Now $a(b-1)=4$ gives $2a=4$ , so $a=2$ . Last, $f(0)=a+c=7$ gives $c=5$ . The exact function is

\boxed{f(x)=2\cdot3^x+5}.

Check it against the last row: $f(2)=2\cdot3^2+5=23$ , which matches.

Try it yourself:

It can try, with a regression that fits the form to the table. But a regression can return decimal estimates, and a decimal won’t always tell you whether $b$ is exactly $3$ or only close to it. Subtracting outputs takes less setup, gives you $b$ , then $a$ , then $c$ as exact numbers, and leaves you a table to check them against.

A hard case the SAT repeats: a function inside a function

In $p(p(x))$ , the output of the inner $p$ becomes the input of the outer $p$ . A function inside a function like this is called a composite.

Going forward takes one step per layer. To evaluate a composite at one number when the arithmetic is quick, work from the inside out by hand, in the same order Desmos followed for $p(p(-2))=p(3)=8$ above.

Working backward to find every input is where answers go missing. The outer equation can have more than one solution, and each of those can come from more than one inner input. That's a lot of branches to track by hand, so let Desmos do the counting. For the same function,

p(x)=x^2-1,

here's how to find every $x$ with $p(p(x))=0$ :

Define $p$ on its own line, so Desmos knows the rule.
On the next line, enter the whole equation, $p(p(x))=0$ . Desmos then graphs every branch at once, instead of one layer at a time.
Select every root.

p(x)=x^2-1
p(p(x))=0

Desmos shows $x\approx-1.414$ , $x=0$ , and $x\approx1.414$ . Seeing all three at once keeps you from losing a branch hidden in the inner function.

Those decimals are enough when they tell the answer choices apart. If the question wants exact values, like $\sqrt2$ instead of $1.414$ , or you need to show where each root comes from, finish with algebra. Give the inner output a temporary name, $u$ :

p(u)=0 \quad\Longrightarrow\quad u^2-1=0 \quad\Longrightarrow\quad u=1\text{ or }u=-1.

Now send both values of $u$ back to the inner function:

p(x)=1 \Longrightarrow x^2-1=1 \Longrightarrow x=\pm\sqrt2,

and

p(x)=-1 \Longrightarrow x^2-1=-1 \Longrightarrow x=0.

So the complete set of real solutions is

\boxed{\{-\sqrt2,\,0,\,\sqrt2\}}.

Check your understanding:

Why would solving only $p(x)=1$ miss a valid input in the example above?

Reading a graph's features, building models from word problems, and shifting graphs come later, in nonlinear graph features, exponential models, and nonlinear function transformations.

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Select all three roots. Each one is an input that makes p(p(x)) equal 0.

Choose hand, Desmos, or both

Ask what the question needs most: every solution, an exact value, or an input the function allows. A graph makes it hard to miss a solution. Algebra keeps values exact and shows you the domain check.

Work by hand when…

the substitution is short, like $p(-2)$ for $p(x)=2x^2-3x+1$ . That's usually faster than typing in the rule.
you can see a shortcut, like the common base in $2^{x+1}=32=2^5$ , a factorization, or outputs one step apart that you can subtract.
the graph has found the solutions and you need their exact values, like $2-\sqrt3$ instead of $0.268$ .
simplifying could hide an input the function doesn't allow. $\dfrac{(x-2)(x+1)}{x-2}$ simplifies to $x+1$ , but $x=2$ is still ruled out.

Use Desmos, or a hybrid with algebra after, when…

a line like $y=5$ may cross the graph more than once, as in the rational example. The graph shows you how many solutions to find.
the rule is long, the input is awkward, or you need the same rule again and again, as in $p(p(-2))$ . Define it once and reuse its name.
you're working backward through a composite, like $p(p(x))=0$ , where one output can lead back to several inputs.
decimals are enough to tell the answer choices apart. If only one choice is close to $1.414$ , you don't need to show it's $\sqrt2$ .

Either way, check every answer against the original function and the question's conditions before you pick a choice.

Practice problems

These run from a direct evaluation to a restricted input, exact parameters, and a function inside a function. Before each one, decide: by hand, Desmos, or both?

Evaluate a rational function

Practice problem

The function $g$ is defined by

g(t)=\frac{2t^3-5}{t+2},

where $t\ne-2$ . What is the value of $g(-1)$ ?

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Plug in by hand, or define g once and enter g(-1).

Keep the denominator's restriction

Practice problem

The function $h$ is defined by

h(x) = \frac{(x-2)(x-3)(x+5)}{x-2},

where $x\ne2$ . What is the sum of all real values of $w$ for which $h(7-w)=0$ ?

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Desmos is fine here, but the original denominator still rules out x = 2.

Find the parameters from outputs one step apart

Practice problem

The function $g$ is defined by $g(x)=ab^x+c$ , where $a\ne0$ , $b>0$ , and $b\ne1$ . Some values of $x$ and $g(x)$ are shown.

$x$	$g(x)$
$0$	$11$
$1$	$17$
$2$	$35$

What is the value of $g(3)$ ?

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Desmos is fine for the arithmetic, but keep a, b, and c exact.

Work backward through a composite

Practice problem

The function $p$ is defined by

p(x)=(x-2)^2-1.

What is the sum of all distinct real solutions to $p(p(x))=0$ ?

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Define p, graph p(p(x))=0 as in the composite example, and count every root before you add.

Finish the lesson

4 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

First ask which way the question goes: input to output, or output back to input.
With a known input, substitute the whole input, wrapped in parentheses, and evaluate.
When a rule is long, repeated, or nested, define it once in Desmos and call it by name.
With a known output, write an equation, find every solution, and test each one against the original function and its domain.
For $ab^x+c$ , subtract outputs one step apart. The changes give you the base first, then the other parameters.
To work backward through a composite, define the function in Desmos, enter the whole equation, and select every root. Add algebra only when you need exact values or have to show where they come from.

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484 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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