2.4.4 Practice set: using derivatives

Maxima and minima

Maxima and minima (lesson 2.4.1)

0 of 7 correct

1.
At what $x$ is $x^2 + 10x$ smallest?
2.
At what $x$ is $xe^{-2x}$ largest?
3.
$f'(x) = (x - 1)(x - 4)$ . At which $x$ does $f$ have a local maximum?
4.
What is the largest value of $x^3 - 12x$ on $[-3, 5]$ ?
5.
What is the smallest value of $x^2$ on $[1, 3]$ ?
6.
How many local maxima and minima does $x^3 - 3x^2 + 3x$ have?
7.
What is the smallest value of $|x + 3| - 2$ ?

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.

Concavity

Concavity and the second derivative (lesson 2.4.2)

0 of 7 correct

1.
Find the second derivative of $x^3 - 3x$ at $x = 2$
2.
Find the second derivative of $e^{-x}$ at $x = 0$
3.
Find the second derivative of $\sqrt{x}$ at $x = 1$
4.
Estimate $f''(1)$ for $f(x) = x^3$ by the second central difference with $h = 0.1$
5.
Where is the inflection point of $x^3 + 3x^2$ ?
6.
$x^3 - 6x^2 + 4$ has critical points at 0 and 4. Use $f''$ to find its local maximum value
7.
$x^4 + 2$ has $f'(0) = 0$ and $f''(0) = 0$ . What is its smallest value?

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 with derivatives (lesson 2.4.3)

0 of 7 correct

1.
For $f(x) = x^2 - 7$ , take one Newton step from $x_0 = 3$
2.
For $f(x) = x^3 - 2$ , take one Newton step from $x_0 = 1$
3.
For $f(x) = \ln x - 1$ , with slope $\frac{1}{x}$ , take one Newton step from $x_0 = 2$
4.
Two Newton errors in a row are 0.01 and 0.0002. With the same quadratic constant, what is the next error?
5.
To minimise $x^2 - 4x$ , Newton's method on $f'(x) = 2x - 4$ , with $f''(x) = 2$ , starts at $x_0 = 10$ . Where does it land?
6.
$f(0) < 0 < f(2)$ , so $[0, 2]$ is a bracket. At $x = 1$ , $f(1) > 0$ and the Newton step would land on 2.5. Where does a bracketed step go instead?
7.
Newton's method on $f'(x) = 3x^2 - 3$ , with $f''(x) = 6x$ , from $x_0 = -2$ lands on $-1$ . Is that a maximum (type 1) or a minimum (type 0) of $x^3 - 3x$ ?

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

Newton's method with a bracket

Write safe_newton(f, fprime, a, b, tol=1e-12, max_steps=100) that returns a root of f in the bracket $[a, b]$ , as in lesson 2.4.3.

If f(a) or f(b) is 0, return that end. If they have the same sign, raise ValueError: there is no bracket.
Start at the midpoint. After each guess $x$ , shrink the bracket to the half that still has the sign change.
Take the Newton step $x - \frac{f(x)}{f'(x)}$ when fprime(x) is not 0 and the step lands strictly inside the bracket. Otherwise take the midpoint of the bracket.
Return when a step moves $x$ by less than tol, or when f(x) is exactly 0. Raise RuntimeError after max_steps steps.

The tests include the cycle of lesson 1.3.3, a flat tangent at the start and the runaway $xe^{-x}$ . One test allows only 12 steps, so bisection alone is too slow.

See the tests

import math


def test_square_root():
    """x² − 2 on [1, 2] gives √2"""
    expect_close(safe_newton(lambda x: x**2 - 2, lambda x: 2 * x, 1, 2), math.sqrt(2), what="root of x² − 2", rel=1e-12)


def test_x_exp_x():
    """xeˣ − 1 on [0, 1] gives about 0.567143"""
    r = safe_newton(lambda x: x * math.exp(x) - 1, lambda x: math.exp(x) * (1 + x), 0, 1)
    expect_close(r, 0.567143290409784, what="root of xeˣ − 1", rel=1e-12)


def test_fast():
    """It takes Newton steps: xeˣ − 1 on [0, 1] converges within 12 steps"""
    try:
        r = safe_newton(lambda x: x * math.exp(x) - 1, lambda x: math.exp(x) * (1 + x), 0, 1, max_steps=12)
    except RuntimeError:
        raise AssertionError("no answer within 12 steps: bisection alone needs about 40, Newton about 6")
    expect_close(r, 0.567143290409784, what="root of xeˣ − 1 within 12 steps", rel=1e-12)


def test_cycle():
    """x³ − 2x + 2 on [−3, 3], where plain Newton from 0 cycles, gives about −1.769292"""
    r = safe_newton(lambda x: x**3 - 2 * x + 2, lambda x: 3 * x**2 - 2, -3, 3)
    expect_close(r, -1.769292354238631, what="root of x³ − 2x + 2", rel=1e-12)


def test_flat_start():
    """x³ − 3x − 1 on [0, 2] starts on a flat tangent at 1 and gives about 1.879385"""
    r = safe_newton(lambda x: x**3 - 3 * x - 1, lambda x: 3 * x**2 - 3, 0, 2)
    expect_close(r, 1.8793852415718169, what="root of x³ − 3x − 1 in [0, 2]", rel=1e-12)


def test_runaway():
    """xe⁻ˣ on [−1, 4], where plain Newton from 1.5 runs away, gives 0"""
    r = safe_newton(lambda x: x * math.exp(-x), lambda x: (1 - x) * math.exp(-x), -1, 4)
    expect_close(r, 0.0, what="root of xe⁻ˣ", abs_tol=1e-12)


def test_root_at_end():
    """A root at an end of the bracket is returned: x² − 4 on [2, 5] gives 2"""
    expect_close(safe_newton(lambda x: x**2 - 4, lambda x: 2 * x, 2, 5), 2.0, what="root of x² − 4 on [2, 5]")


def test_no_bracket():
    """x² + 1 on [−1, 1] has no sign change: ValueError"""
    try:
        safe_newton(lambda x: x**2 + 1, lambda x: 2 * x, -1, 1)
    except ValueError:
        return
    raise AssertionError("expected ValueError when f(a) and f(b) have the same sign")


def test_max_steps():
    """Stopping after max_steps without converging raises RuntimeError"""
    try:
        safe_newton(lambda x: x**2 - 2, lambda x: 2 * x, 1, 2, max_steps=2)
    except RuntimeError:
        return
    raise AssertionError("expected RuntimeError after 2 steps without converging")

Maxima and minima

Concavity

Newton's method

Coding exercise

Free further reading