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Function Notation and Evaluation: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 An input fraction

Let $f(t)=t(3-t)$ for real $t$ . Evaluate $f(\frac12)$ .
1. Hint 1
  The full fraction goes into every occurrence of the argument.
2. Hint 2
  Compute the difference inside the parentheses before multiplying.
Answer

$f(\frac12)=\frac54$ .

Full solution

Substitute the same input into both slots.
$\begin{aligned}f(\tfrac12)&=\tfrac12(3-\tfrac12)\\&=\tfrac12\cdot\tfrac52\\&=\tfrac54\end{aligned}$

Answer

$f(\frac12)=\frac54$ .

Key idea

Every occurrence of a function argument receives the same complete input.
Problem 1: Mark complete
Problem 2 A selected branch

A function on $\mathbb R$ is given by $p(t)=2-t$ when $t\le-1$ , and $p(t)=t(t+1)$ when $t>-1$ . Find $p(-1)$ .
1. Hint 1
  The branch condition is decided before evaluating its formula.
2. Hint 2
  The equality sign tells which rule includes the boundary input.
Answer

$p(-1)=3$ .

Full solution

The input $-1$ satisfies $t\le-1$ , so use the first rule.
$\begin{aligned}p(-1)&=2-(-1)\\&=3\end{aligned}$

The second branch excludes $-1$ and contributes no alternative output.

Answer

$p(-1)=3$ .

Key idea

A piecewise boundary belongs to the case whose inequality includes it.
Problem 2: Mark complete
Problem 3 A complete replacement

Let $f(t)=t(2-t)$ on $\mathbb R$ . Write $f(1+u)$ as a polynomial in real $u$ .
1. Hint 1
  The input expression must replace both occurrences of $t$ .
2. Hint 2
  After substituting, simplify $2-(1+u)$ before multiplying.
Answer

$f(1+u)=1-u^2$ .

Full solution

The substitution gives
$\begin{aligned}f(1+u)&=(1+u)(1-u)\\&=1-u^2\end{aligned}$

The constant input $u=0$ gives $f(1)=1$ , agreeing with the resulting polynomial.

Answer

$f(1+u)=1-u^2$ .

Key idea

Parentheses keep a compound input intact in every slot of a rule.
Problem 3: Mark complete
Problem 4 A parking charge

A parking lot charges $C(t)=4$ dollars for real times $0<t\le1$ hour and $C(t)=4+3(t-1)$ dollars for $1<t\le5$ hours. Find the increase in charge from $t=\frac12$ hour to $t=\frac52$ hours.
1. Hint 1
  The two times lie in different branches of the charge rule.
2. Hint 2
  Evaluate each branch at its own input, then subtract the earlier charge from the later one.
Answer

An increase of $\frac92$ dollars, or $4.50$ dollars.

Full solution

At $t=\frac12$ , the charge is $4$ dollars.
At $t=\frac52$ ,
$\begin{aligned}C(\tfrac52)&=4+3(\tfrac52-1)\\&=\tfrac{17}{2}\end{aligned}$

The increase is $\frac{17}{2}-4=\frac92$ dollars.
The original time inputs are within the declared domain.

Answer

An increase of $\frac92$ dollars, or $4.50$ dollars.

Key idea

Evaluate each input in its own branch before comparing function values.
Problem 4: Mark complete
Problem 5 Two nearby inputs

Let $f(x)=x(4-x)$ for all real $x$ . For real $h\ne0$ , simplify $\frac{f(x+h)-f(x)}h$ and state what it measures on the graph.
1. Hint 1
  The numerator is the change in output between two distinct inputs.
2. Hint 2
  Expand $(x+h)(4-x-h)$ , subtract $x(4-x)$ , and factor out $h$ .
Answer

$4-2x-h$ , for $h\ne0$ ; the secant slope from $(x,f(x))$ to $(x+h,f(x+h))$ .

Full solution

Expand the shifted output:
$f(x+h)=(x+h)(4-x-h).$
Multiplying out gives
$4x-x^2-2xh+4h-h^2$
Subtracting $f(x)=4x-x^2$ leaves $4h-2xh-h^2$ , which factors, since $h$ divides every term, as
$h(4-2x-h)$

Dividing by the nonzero $h$ gives $4-2x-h$ .
The horizontal change between the graph points is $h$ and the vertical change is the numerator, so the quotient is the secant slope, or average rate of change.

Answer

$4-2x-h$ , for $h\ne0$ ; the secant slope from $(x,f(x))$ to $(x+h,f(x+h))$ .

Key idea

A difference quotient measures output change per unit input change between two distinct points.
Problem 5: Mark complete
Problem 6 A hidden input name

A function $F$ on $\mathbb R$ satisfies $F(2t-1)=t^2-t$ for every real $t$ . Find a formula for $F(u)$ as a polynomial in $u$ .
1. Hint 1
  The slot inside $F$ holds the whole expression $2t-1$ , not the bare variable $t$ ; find the value of $t$ that makes that expression equal an arbitrary $u$ .
2. Hint 2
  Set $u=2t-1$ and solve for $t$ , then substitute that expression into $t^2-t$ .
Answer

$F(u)=\frac{u^2-1}{4}$ .

Full solution

Set $u=2t-1$ , so
$t=\frac{u+1}{2}.$
Substituting yields
$\begin{aligned}F(u)&=\frac{(u+1)^2}{4}-\frac{u+1}{2}\\&=\frac{u^2-1}{4}\end{aligned}$

Every real $u$ comes from exactly one real $t$ , so this defines the formula on the full domain.
Replacing $u$ with $2t-1$ returns $t^2-t$ .

Answer

$F(u)=\frac{u^2-1}{4}$ .

Key idea

Where each value of the compound input comes from exactly one value of the original variable, solve for that variable and substitute back to recover the rule.
Problem 6: Mark complete
Problem 7 A naming decision

Two devices define functions on $D=\{-1,0,1\}$ . The first uses $f(t)=t^2+t$ with codomain $\{0,2\}$ ; the second uses $g(u)=u(u+1)$ with codomain $\{0,1,2\}$ . Decide whether they are equal functions, and give one change to the second device data that would make them equal.
1. Hint 1
  Equality involves the declared sets as well as the computed outputs.
2. Hint 2
  Expand the second rule and compare the two codomains.
Answer

They are different; change the second codomain to $\{0,2\}$ .

Full solution

The rules give the same outputs since
$u(u+1)=u^2+u.$
Both domains are $D$ , but their declared codomains differ.

On $D$ , the output set is $\{0,2\}$ , so replacing the second codomain by $\{0,2\}$ is valid and makes the domains, codomains, and all outputs match.

Answer

They are different; change the second codomain to $\{0,2\}$ .

Key idea

Equivalent formulas and matching domains still need matching codomains for equality of functions.
Problem 7: Mark complete
Problem 8 A label on the machine

A display labels a rule $f$ and shows $f(3)=11$ . A student erases the label $f$ and replaces it with $11$ , saying the function and its displayed output are interchangeable. Explain why that is not justified, giving two different functions on $\mathbb R$ with that displayed output.
1. Hint 1
  One output records what happened at one input, not the whole rule.
2. Hint 2
  Choose two rules that agree at $3$ but differ at another real input.
Answer

Not justified; for example $f(x)=x+8$ and $g(x)=2x+5$ both give $11$ at $3$ and differ elsewhere.

Full solution

The symbol $f$ names the whole function; $f(3)$ names one value.
For example,
$\begin{aligned}3+8&=11\\2(3)+5&=11\end{aligned}$

But the two proposed rules give $8$ and $5$ at input $0$ .
Thus the displayed value does not identify the function and cannot replace its name.

Answer

Not justified; for example $f(x)=x+8$ and $g(x)=2x+5$ both give $11$ at $3$ and differ elsewhere.

Key idea

A function value at one input does not determine the function itself.
Problem 8: Mark complete
Problem 9 A claim about combining inputs

Let $f(x)=\lvert x\rvert$ for real $x$ . A student claims $f(a+b)=f(a)+f(b)$ for all real $a,b$ . Assess the claim with one pair for which it fails and one pair for which it holds.
1. Hint 1
  Combining inputs happens before the magnitude is taken.
2. Hint 2
  Compare inputs with opposite signs, then inputs that are both positive.
Answer

False in general, for example: $(a,b)=(2,-2)$ gives $0\ne4$ ; $(2,3)$ gives $5=5$ .

Full solution

For $a=2,b=-2$ ,
$\begin{aligned}f(a+b)&=0\\f(a)+f(b)&=4\end{aligned}$
so the universal claim fails.

For $a=2,b=3$ , the combined input has magnitude $5$ , matching $2+3=5$ .
An identity can hold for particular inputs without holding for every pair.

Answer

False in general, for example: $(a,b)=(2,-2)$ gives $0\ne4$ ; $(2,3)$ gives $5=5$ .

Key idea

An equality for selected inputs is weaker than an identity that holds for all inputs.
Problem 9: Mark complete
Problem 10 A scaled argument

For $f(x)=-\frac32x$ on $\mathbb R$ , a student claims $f(kx)=kf(x)$ for every real $k,x$ . Is the claim correct? Prove the verdict directly from the rule.
1. Hint 1
  Track where the factor $k$ enters the rule.
2. Hint 2
  Compare the two products using associativity and commutativity of multiplication.
Answer

Yes; $f(kx)=kf(x)$ for every real $k,x$ .

Full solution

Substitution gives
$f(kx)=-\frac32(kx).$
Associativity and commutativity rewrite this as $k(-\frac32x)$ , which is $kf(x)$ .

Both expressions are defined for every real $k,x$ , so the equality holds throughout the stated domain.

Answer

Yes; $f(kx)=kf(x)$ for every real $k,x$ .

Key idea

A constant multiple of the input respects scaling because real multiplication is associative and commutative.
Problem 10: Mark complete

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