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Radical Functions: 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: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra I. You can skip it.

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Problem 1 of 10

Problem 1 The allowed inputs

Find the domain of $f(x)=\sqrt{7-2x}+1$ .
1. Hint 1
  Here only what sits under the even root can restrict the inputs, so look at the radicand alone.
2. Hint 2
  Set the radicand to be nonnegative and solve for $x$ , checking the inequality direction once $x$ is isolated.
Answer

$x\le\tfrac72$ , that is $x\le3.5$ .

Full solution

The $1$ is added outside the root, so it places no restriction on the input.
The root has even index, so its radicand must be nonnegative, which requires
$7-2x\ge0$

Adding $2x$ to both sides gives $7\ge2x$ , and dividing by the positive $2$ gives
$x\le\tfrac72$
So the accepted inputs are every $x\le3.5$ .

At $x=3.5$ the radicand is $0$ , so $f(3.5)=1$ .
At $x=4$ the radicand is $-1$ , which no real number squares to, so that input is excluded.

Answer

$x\le\tfrac72$ , that is $x\le3.5$ .

Key idea

A constant added outside an even root leaves the domain to the sign condition on the radicand.
Problem 1: Mark complete
Problem 2 The machine setting

A machine can take either the fourth root or the fifth root of its input. It must return a real number for every real input. Which root index should be installed?
1. Hint 1
  The setting must work for negative inputs as well as zero and positive inputs.
2. Hint 2
  Compare which real values an even power and an odd power can produce.
Answer

$5$ .

Full solution

A fourth power is never negative, so a fourth-root setting cannot return a real number for a negative input.
For example, no real number has fourth power $-1$ .

A fifth power can produce every real value, with negative, zero, and positive outputs.
Its inverse, the fifth root, therefore accepts every real input, so install index $5$ .

Answer

$5$ .

Key idea

An odd-index root accepts every real input because the corresponding odd power can produce every real output.
Problem 2: Mark complete
Problem 3 The enclosed square

Find the real domain of $f(x)=\sqrt{(x-2)^2}$ .
1. Hint 1
  The even root needs its entire radicand to be zero or positive.
2. Hint 2
  Test a few inputs on either side of $x=2$ , including $x=2$ itself, and decide what the radicand can never be.
Answer

All real numbers.

Full solution

For every real input,
$(x-2)^2\ge0$
Thus every input meets the square root restriction.

In particular, $x=2$ gives radicand zero and is included; inputs on both sides give positive radicands.

Answer

All real numbers.

Key idea

A squared real expression already satisfies the sign restriction of a square-root radicand.
Problem 3: Mark complete
Problem 4 The two curves

On the axes in the figure, sketch $y=\sqrt{x}$ and $y=x/2$ for $0\le x\le4$ . Find their meeting points and determine which graph is higher between them.

Blank axes for your sketch.

Text description of this figure

Blank coordinate axes for your own sketch. The horizontal axis is labeled x and runs from negative one to five; the vertical axis is labeled y and runs from negative one to four. Both axes use the same unit length, with a light square grid, tick marks and number labels at every whole number, the origin labeled 0, and arrowheads at both ends of each axis. Nothing at all is plotted on the grid.
1. Hint 1
  The root curve starts at zero and rises more slowly as the input grows.
2. Hint 2
  At a meeting point, both nonnegative outputs agree, so their squares agree.
Answer

The root curve through $(0,0)$ , $(1,1)$ and $(4,2)$ , bending flatter as it rises; the line is the segment from $(0,0)$ to $(4,2)$ ; meeting points $(0,0)$ and $(4,2)$ ; the root curve is higher for $0<x<4$ .

Full solution

The root curve passes through $(0,0)$ , $(1,1)$ , and $(4,2)$ and bends downward as it rises.
The line is the segment joining $(0,0)$ to $(4,2)$ .

At an intersection, squaring $\sqrt{x}=x/2$ is valid because both sides are nonnegative.
It gives
$x^2=4x$
Hence $x(x-4)=0$ , giving the two stated points, both of which check.

For $0<x<4$ , multiplication by $x>0$ gives $x^2<4x$ .
Thus $(x/2)^2<x$ , so the nonnegative quantity $x/2$ is smaller than $\sqrt{x}$ .

Answer

The root curve through $(0,0)$ , $(1,1)$ and $(4,2)$ , bending flatter as it rises; the line is the segment from $(0,0)$ to $(4,2)$ ; meeting points $(0,0)$ and $(4,2)$ ; the root curve is higher for $0<x<4$ .

Key idea

Comparing squares preserves the order of nonnegative outputs and can locate one graph relative to another.
Problem 4: Mark complete
Problem 5 The marked endpoint

The figure shows a function of the form $f(x)=a\sqrt{x-h}+k$ , with $a\ne0$ . Write its rule and state its domain and range.

The graph of $f$ , with two points marked.

Text description of this figure

A coordinate grid with equal unit spacing on both axes. The horizontal axis is labeled x and numbered from negative five to six; the vertical axis is labeled y and numbered from negative six to two, and the origin is labeled 0. A curve labeled f begins at a filled dot at the point (negative four, one), which is its leftmost point, and falls to the right, steeply at first and then more and more gently. It passes through a second filled dot at the point (0, negative three) and keeps falling to the right edge of the grid, where an arrow shows that it continues. Those two points are the only ones labeled, and no rule, equation or other value is printed.
1. Hint 1
  The starting point supplies both shifts.
2. Hint 2
  Use the other marked point to determine the vertical multiplier.
Answer

$f(x)=-2\sqrt{x+4}+1$ ; domain $x\ge-4$ ; range $y\le1$ .

Full solution

The starting point $(-4,1)$ gives $h=-4$ and $k=1$ .
Substituting the marked point $(0,-3)$ makes the radicand $0+4=4$ , whose root is $2$ , so
$-3=2a+1$
and $a=-2$ .

The rule is
$f(x)=-2\sqrt{x+4}+1$
Its radicand requires $x\ge-4$ , and its negative multiplier sends every root value downward from height $1$ , giving $y\le1$ .

At $x=0$ , the rule returns $-3$ , checking the marked point.

Answer

$f(x)=-2\sqrt{x+4}+1$ ; domain $x\ge-4$ ; range $y\le1$ .

Key idea

A radical graph can be reconstructed from its endpoint and one further point.
Problem 5: Mark complete
Problem 6 The nested machine

A machine first computes $u=\sqrt{x-3}$ and then reports $\sqrt{9-u}$ . Find all real inputs the machine accepts, and the machine's output at each extreme input it accepts.
1. Hint 1
  Each square root must receive a nonnegative radicand.
2. Hint 2
  After requiring $x\ge3$ , require the first output to be at most $9$ .
Answer

$3\le x\le84$ ; output $3$ at $x=3$ and $0$ at $x=84$ .

Full solution

The first root requires $x\ge3$ .
The second requires
$\sqrt{x-3}\le9.$
Both quantities are nonnegative, so squaring gives
$x-3\le81$
Therefore the accepted inputs are
$3\le x\le84.$

At $x=3$ , the intermediate output is $0$ and the final output is $\sqrt9=3$ .
At $x=84$ , the intermediate output is $9$ and the final output is $\sqrt0=0$ .
Both endpoints are allowed.

Answer

$3\le x\le84$ ; output $3$ at $x=3$ and $0$ at $x=84$ .

Key idea

Composed radical operations must satisfy the domain restriction at every stage.
Problem 6: Mark complete
Problem 7 The square panel

A square panel has area $12-x$ square cm and side length $x-6$ cm, where $x$ is real and the side is positive. Find $x$ and the panel dimensions, checking every algebraic candidate.
1. Hint 1
  The nonnegative square root of the area must equal the positive side.
2. Hint 2
  Squaring gives a quadratic, but a candidate with a negative stated side must be rejected.
Answer

$x=8$ ; side $2$ cm; area $4$ square cm; the candidate $x=3$ is rejected.

Full solution

The conditions include $x>6$ and $12-x\ge0$ .
The side equation is
$\sqrt{12-x}=x-6$
Squaring and collecting gives
$x^2-11x+24=0$

Factoring gives $(x-3)(x-8)=0$ , so the candidates are $3$ and $8$ .
At $3$ , the principal root is $3$ but the stated side is $-3$ , so that candidate fails.

At $8$ , the area is $4$ square cm and the side is $2$ cm.
The principal root is $2$ , so this candidate checks and gives the requested dimensions.

Answer

$x=8$ ; side $2$ cm; area $4$ square cm; the candidate $x=3$ is rejected.

Key idea

A candidate from squaring survives only if it satisfies the original equation and every stated condition, here that the side $x-6$ is positive.
Problem 7: Mark complete
Problem 8 The combined readings

Mina claims $\sqrt{u^2}+\sqrt{v^2}=\sqrt{(u+v)^2}$ for every pair of real numbers $u,v$ . Is she correct? Justify your decision.
1. Hint 1
  Each root reports a nonnegative size, while the sum inside the right root retains signs before squaring.
2. Hint 2
  Compare what each side reports when $u$ and $v$ share a sign with what it reports when their signs differ.
Answer

No; $u=2$ , $v=-2$ give left side $4$ and right side $0$ .

Full solution

For $u=2$ and $v=-2$ , the two separate roots each equal $2$ , so their sum is
$2+2=4$

But $u+v=0$ , so the right side is $\sqrt0=0$ .
The unequal results disprove the claim.

Answer

No; $u=2$ , $v=-2$ give left side $4$ and right side $0$ .

Key idea

Combining signed inputs before taking a principal root can give a different result from adding their separate nonnegative sizes.
Problem 8: Mark complete
Problem 9 The two root records

For $x\ge9$ , consider $\sqrt{x-4}=\sqrt{x-9}$ . Ivo says squaring this equation cannot introduce an extraneous candidate because both sides are nonnegative. Is he correct, and does the equation have any solutions?
1. Hint 1
  On the stated domain both roots exist and are nonnegative.
2. Hint 2
  Both radicands are already isolated under single roots, so compare what the squared equation says about $x-4$ and $x-9$ .
Answer

Ivo is correct; there are no solutions.

Full solution

On $x\ge9$ , both sides are nonnegative.
Squaring is reversible for such values, so it introduces no extraneous candidate here.

The squared equation is
$x-4=x-9$
Subtracting $x$ would require $-4=-9$ , which is false.
Thus no candidate exists, and the original equation has no solutions.

Answer

Ivo is correct; there are no solutions.

Key idea

Squaring an equality is reversible when both sides are known to be nonnegative.
Problem 9: Mark complete
Problem 10 The candidate list

Choose a real constant $c$ so that squaring $\sqrt{x+c}=x$ produces candidates $x=-1$ and $x=2$ . Find $c$ and decide which candidates satisfy the original equation.
1. Hint 1
  The squared equation is a quadratic whose two roots are specified.
2. Hint 2
  Build that quadratic from the two root factors, then compare it with $x^2-x-c=0$ .
Answer

$c=2$ ; $x=2$ is valid and $x=-1$ is extraneous.

Full solution

The prescribed roots give
$(x+1)(x-2)=x^2-x-2$
Comparing with $x^2-x-c$ gives $c=2$ .

At $x=2$ , $\sqrt{2+2}=2$ , so the equation holds.
At $x=-1$ , the left side is $\sqrt1=1$ , not $-1$ , so that candidate is extraneous.

Answer

$c=2$ ; $x=2$ is valid and $x=-1$ is extraneous.

Key idea

Reconstructing the squared equation does not replace checking the signs in the original radical equation.
Problem 10: Mark complete

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