1.3.4 Practice set: roots

Polynomials

Polynomials and their roots (lesson 1.3.1)

0 of 6 correct

1.
Evaluate $x^3 - 6x^2 + 11x - 6$ at $x = 4$
2.
Evaluate $x^3 - 4x$ at $x = -3$
3.
$x^2 + 2x - 15 = 0$ has two roots. What is the larger one?
4.
2 is a root of $x^3 - 3x^2 + 4$ , so it equals $(x - 2)(x^2 + bx + c)$ . What is $c$ ?
5.
$x^3 - 3x^2 + 4 = (x - 2)^2(x + 1)$ . At which root does the graph touch the axis without crossing?
6.
A polynomial of degree 5 must have at least how many real roots?

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

Bisection

Bisection (lesson 1.3.2)

0 of 6 correct

1.
$p(1) = -2$ and $p(3) = 4$ , so $[1, 3]$ is a bracket. What is its first midpoint?
2.
For $f(x) = x^2 - 3$ on $[1, 2]$ : $f(1.5) = -0.75$ . The root lies in $[1.5, b]$ : what is $b$ ?
3.
For the falling function $5 - x^2$ on $[2, 3]$ : $f(2) = 1$ and $f(2.5) = -1.25$ . The root lies in $[a, 2.5]$ : what is $a$ ?
4.
After 5 halvings, how wide is a bracket that started with width 4?
5.
How many halvings shrink a bracket of width 4 below $10^{-6}$ ?
6.
$\frac{1}{x}$ is $-1$ at $x = -1$ and $\frac{1}{2}$ at $x = 2$ . Bisection on $[-1, 2]$ closes in on which $x$ ?

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

Newton's method

Newton's method (lesson 1.3.3)

0 of 6 correct

1.
For $f(x) = x^2 - 10$ , with slope $2x$ , take one Newton step from $x_0 = 3$
2.
For $f(x) = x^3 - 2$ , with slope $3x^2$ , take one Newton step from $x_0 = 1$
3.
For $f(x) = x^2 - 4$ , with slope $2x$ , Newton from $x_0 = 1$ gives $x_1 = 2.5$ . What is $x_2$ ?
4.
For $f(x) = 2x - 6$ (slope 2), Newton from any start lands on the root in one step. What is it?
5.
For $x^3 - 2x + 2$ , with slope $3x^2 - 2$ , Newton from $x_0 = 0$ gives $x_1 = 1$ . What is $x_2$ ?
6.
$x^2 - 4x + 5$ has slope $2x - 4$ . From which start does Newton's method fail at once?

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

Coding exercise

Exercise

Bisection, with a safety check

Write bisect(f, a, b, tol=1e-10) that returns a root of f between a and b, accurate to within tol.

If f(a) and f(b) have the same sign, the bracket proves nothing: there may be no root, two roots or a double root inside. Raise ValueError.
If f(a) or f(b) is already exactly 0, return that end.
Otherwise halve the interval until it is narrower than tol, keeping the half with the sign change, and return its midpoint.

The tests use rising and falling functions, a bracket with three roots inside, and brackets with no sign change.

See the tests

import math


def test_square_root_of_two():
    """x² − 2 on [1, 2] gives √2"""
    expect_close(bisect(lambda x: x**2 - 2, 1, 2), math.sqrt(2), what="bisect(x² − 2, 1, 2)", rel=1e-9)


def test_cubic():
    """x³ − x − 1 on [1, 2]: the answer makes the cubic zero"""
    r = bisect(lambda x: x**3 - x - 1, 1, 2)
    expect_close(r**3 - r - 1, 0.0, what="f at your root", abs_tol=1e-8)


def test_decreasing_function():
    """A falling function works too: 5 − x on [0, 10] gives 5"""
    expect_close(bisect(lambda x: 5 - x, 0, 10), 5.0, what="bisect(5 − x, 0, 10)", rel=1e-9)


def test_tolerance():
    """A looser tolerance still lands within it: x² − 2 with tol = 0.01"""
    r = bisect(lambda x: x**2 - 2, 1, 2, tol=0.01)
    assert abs(r - math.sqrt(2)) < 0.01, f"expected within 0.01 of √2, got {r}"


def test_root_at_an_end():
    """If f(a) is already 0, return a"""
    expect_close(bisect(lambda x: x - 1, 1, 3), 1.0, what="bisect(x − 1, 1, 3)")


def test_no_bracket():
    """No sign change between a and b raises ValueError"""
    try:
        bisect(lambda x: x**2 + 1, -1, 1)
    except ValueError:
        return
    raise AssertionError("expected ValueError: x² + 1 is positive at both ends")


def test_root_at_b():
    """If f(b) is already 0, return b"""
    expect_close(bisect(lambda x: x - 3, 1, 3), 3.0, what="bisect(x − 3, 1, 3)")


def test_several_roots():
    """(x − 1)(x − 2)(x − 3) on [0, 4.5] holds three roots: any one of them is right"""
    r = bisect(lambda x: (x - 1) * (x - 2) * (x - 3), 0, 4.5)
    nearest = min(abs(r - root) for root in (1, 2, 3))
    assert nearest < 1e-9, f"expected one of 1, 2 or 3, got {r}"


def test_double_root():
    """(x − 1)² on [0, 3] has a double root but no sign change: raise ValueError"""
    try:
        bisect(lambda x: (x - 1) ** 2, 0, 3)
    except ValueError:
        return
    raise AssertionError("expected ValueError: (x − 1)² is positive at both ends")

Polynomials

Bisection

Newton's method

Coding exercise

Free further reading