You are handed a formula — say the profit of a product as a function of its price, or the height of a projectile as a function of time — and asked: where is it largest? Plotting helps, but a plot is only as good as the points you sampled, and it never proves anything. You want a way to read the shape of a graph directly from the formula.

The derivative is exactly that tool. Recall that \(f'(x)\) is the slope of the tangent line at \(x\). Slope has a sign, and the sign is a direction:

Walk along a hilly road from left to right. While you climb, your altimeter’s rate of change is positive. At the very top of a hill — for one instant — you are neither climbing nor descending: the rate is zero. Then you start to descend and the rate turns negative. So the summit is betrayed by the rate passing through zero and changing sign from + to −. A valley floor is the reverse: − then 0 then +.

That is the whole idea of this section. Peaks and troughs hide at the places where \(f'\) is zero (or fails to exist), and the sign change of \(f'\) around those places tells you which is which.

Why must a smooth function with a positive derivative everywhere on an interval actually be increasing there? Positive slope at every single point feels like it should force the function up, but "slope at a point" is a limit — an instantaneous statement — and "increasing" compares two different points.

The bridge is the Mean Value Theorem: if \(f\) is differentiable on an interval and \(a < b\) are two points in it, then somewhere between them there is a point \(c\) where the tangent slope equals the average slope:

Picture it: draw the chord (secant) from \((a, f(a))\) to \((b, f(b))\); slide a parallel line toward the curve, and it must touch the curve somewhere as a tangent. Now rearrange: \(f(b) - f(a) = f'(c)\,(b-a)\). If \(f'\) is positive everywhere and \(b - a > 0\), the right-hand side is positive, so \(f(b) > f(a)\). The function is increasing. Same argument with signs flipped for decreasing, and with \(f' = 0\) everywhere it says \(f(b) = f(a)\): a function with zero derivative is constant. (That last fact is the reason antiderivatives come with a "+ C" in Chapter 19.)

The Mean Value Theorem is also the legal basis of average-speed cameras: if your average speed between two cameras was 62 mph, the theorem guarantees that at some instant your speedometer read exactly 62 mph. Over a 50 mph limit, that is enough for a ticket. Try it: Mean value point for √x.

The UK/US clash in the concavity vocabulary is real and causes confusion: a US textbook’s "concave up" is a British "convex". In optimisation and machine learning the British/European usage wins: a convex function is cup-shaped, and convexity guarantees that any local minimum is a global one.

The second derivative gives a quicker test at a stationary point \(c\) (where \(f'(c) = 0\)):

Find and classify the critical points of \(f(x) = x^3 - 3x^2 - 9x + 5\), find where it is concave up and down, and locate its inflection point. Try it: Differentiate the cubic.

Step 1 — differentiate and factor.

Step 2 — critical points. \(f'(x) = 0\) when \(x = -1\) or \(x = 3\). \(f'\) is a polynomial, defined everywhere, so there are no others.

Step 3 — sign chart. The critical points split the line into three intervals. Test one point in each:

Step 4 — classify and evaluate.

Step 5 — check with the second derivative. \(f''(x) = 6x - 6\).

Step 6 — concavity and inflection. \(f''(x) = 6(x - 1)\) is negative for \(x < 1\) and positive for \(x > 1\). The concavity changes at \(x = 1\), where \(f(1) = 1 - 3 - 9 + 5 = -6\). The inflection point is \((1, -6)\) — exactly halfway between the two turning points, as it always is for a cubic.

Sketch \(f(x) = x^4 - 4x^3\).

Notice what the procedure caught that random sampling might miss: the origin is flat but not a turning point. Point-sampling plots tend to show it as a slight kink or not at all.

"Optimisation" is mathematics for the question every engineer asks: what is the best choice? The cheapest design, the shortest route, the highest throughput, the smallest error. In the single-variable form it reads: find the value of \(x\) in some allowed set that makes a quantity \(f(x)\) as large (or as small) as possible.

A smooth function on a closed interval \(\left[a, b\right]\) has to achieve both a highest and a lowest value somewhere — this is the Extreme Value Theorem. The highest point can only be in one of two kinds of place:

Why horizontal? If \(f'(c) > 0\) at an interior point you could step slightly right and go higher; if \(f'(c) < 0\) you could step slightly left and go higher. Either way \(c\) would not be the top. Only \(f'(c) = 0\) (or a point where \(f'\) fails to exist) survives. This is Fermat’s theorem on stationary points, and it reduces an infinite search over every \(x\) to a short list of candidates.

A square sheet of card is 20 cm on each side. You cut an equal square of side \(x\) from each corner and fold up the flaps to make an open-topped box. What \(x\) gives the largest volume? Try it: Design the open box; Graph the box volume.

Picture and names. After cutting, the base is a square of side \(20 - 2x\) (you lose \(x\) at each end), and the height is \(x\).

Objective. Volume = base area × height:

There was only one free variable to begin with, so no constraint is needed to eliminate anything.

Domain. \(x\) must be positive, and \(20 - 2x\) must be positive, so \(0 \le x \le 10\) (allowing the degenerate endpoint boxes, which have zero volume).

Differentiate. Use the product rule with \(u = x\), \(v = (20 - 2x)^2\), so \(u' = 1\) and, by the chain rule, \(v' = 2(20 - 2x)\cdot(-2) = -4(20 - 2x)\):

Critical points. \(V'(x) = 0\) when \(20 - 2x = 0\) (\(x = 10\), an endpoint) or \(20 - 6x = 0\) (\(x = 10/3 \approx 3.333\)).

Compare candidates.

Answer. Cut squares of side \(10/3 \approx 3.33\) cm. The box is \(13.33 \times 13.33 \times 3.33\) cm and holds about \(593\ \text{cm}^3\).

A farmer has 200 m of fencing to enclose a rectangular field along a straight river. No fence is needed along the river. What dimensions maximise the area?

Names. Let \(x\) be the length of each of the two sides perpendicular to the river and \(y\) the side parallel to it.

Objective: \(A = xy\). Constraint: fence used \(= 2x + y = 200\).

Eliminate. \(y = 200 - 2x\), so

Differentiate and solve.

Check. \(A''(x) = -4 < 0\), so concave down everywhere: \(x = 50\) is the global maximum. (Endpoints give \(A(0) = A(100) = 0\).)

Answer. \(x = 50\) m, \(y = 200 - 100 = 100\) m, area \(5000\ \text{m}^2\). The side along the river is twice each of the other sides.

Alternative approach. \(A(x) = 200x - 2x^2\) is a downward parabola; completing the square (Chapter 8) gives \(A = -2(x - 50)^2 + 5000\), so the vertex is at \(x = 50\) — no calculus needed. For quadratics, the vertex formula is the quicker method. Calculus earns its keep when the objective is not a quadratic, as in the next example.

A drinks company wants a closed cylindrical can holding \(500\ \text{cm}^3\). Metal costs the same per square centimetre everywhere, so minimising cost means minimising surface area. What radius and height should they use? Try it: Find the cheapest can.

Names. Radius \(r\), height \(h\), both in cm.

Objective. Surface area = two circular ends + the curved side (which unrolls into a rectangle of width \(2\pi r\), height \(h\)):

Constraint. \(\pi r^2 h = 500\), so \(h = \dfrac{500}{\pi r^2}\).

Eliminate.

Differentiate. Write \(1000/r = 1000 r^{-1}\):

Check. \(S''(r) = 4\pi + 2000/r^3 > 0\) for all \(r > 0\): concave up on the whole domain, so this single critical point is the global minimum. (The domain \(r > 0\) is open — there are no endpoints to check — but as \(r \to 0^+\) or \(r \to \infty\), \(S \to \infty\), confirming a minimum in between.)

Height. Here is the elegant part. From \(r^3 = 250/\pi\) we get \(\pi r^3 = 250\), so

Answer. \(r \approx 4.30\) cm, \(h \approx 8.60\) cm, surface area \(S \approx 348.7\ \text{cm}^2\). The cheapest can is exactly as tall as it is wide. (Real drinks cans are taller and thinner, because thicker metal is used for the ends, the ends are more expensive to form, and a tall can is easier to hold — change the cost per unit area of the ends in exercise 6 and watch the optimum move.)

Choose the lightest tool that works. If the objective is a quadratic, don’t differentiate it twice and run the second-derivative test; just find the vertex.

Air is pumped into a spherical balloon at a steady \(100\ \text{cm}^3\) per second. How fast is the radius growing when the radius is 10 cm? You know one rate (volume per second) and want another (radius per second). The two quantities are tied together by a formula, \(V = \tfrac{4}{3}\pi r^3\), and that link forces their rates to be tied together too.

When several quantities all change with time, each is a function of \(t\). If an equation relates them at every instant, you can differentiate both sides of the equation with respect to \(t\) — the two sides are equal functions of time, so their rates of change are equal. The chain rule (Chapter 17) does the work: the derivative of \(r^3\) with respect to time is \(3r^2 \cdot \frac{dr}{dt}\), "the rate \(V\) changes per unit of \(r\), times the rate \(r\) changes per unit of time". This is implicit differentiation with time as the hidden variable.

Known: \(\frac{dV}{dt} = 100\ \text{cm}^3/\text{s}\). Wanted: \(\frac{dr}{dt}\) when \(r = 10\) cm. Link: \(V = \tfrac{4}{3}\pi r^3\).

Notice that \(4\pi r^2\) is the balloon’s surface area: the new air is spread as a thin shell over the whole surface, so the bigger the balloon, the slower the radius grows.

A 5 m ladder leans against a vertical wall. Its foot slides away from the wall at \(0.5\) m/s. How fast is the top sliding down the wall when the foot is 3 m from the wall?

Names. \(x(t)\): distance of the foot from the wall; \(y(t)\): height of the top. Known: \(\frac{dx}{dt} = 0.5\). Wanted: \(\frac{dy}{dt}\) when \(x = 3\). Link (Pythagoras, true at all times): \(x^2 + y^2 = 25\).

Snapshot: when \(x = 3\), \(y = \sqrt{25 - 9} = 4\).

The top slides down (negative sign) at \(0.375\) m/s. As the top approaches the ground (\(y \to 0\)), \(\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}\) grows without bound — the model breaks down because in reality the ladder’s top leaves the wall.

Without a calculator, what is \(\sqrt{4.1}\)? You know \(\sqrt{4} = 2\) exactly, and \(4.1\) is very close to 4. Surely \(\sqrt{4.1}\) is just a little more than 2 — but how much more?

Zoom in on any smooth curve far enough and it looks straight — it looks like its tangent line. So near a point \(a\) where you know everything, replace the curve by its tangent line and evaluate the line instead. Lines are trivial to evaluate.

The tangent line at \(x = a\) passes through \((a, f(a))\) with slope \(f'(a)\). Point-slope form (Chapter 5) gives

This \(L\) is the linearisation of \(f\) at \(a\). Because \(f'(a)\) is the limit of the slopes \(\frac{f(x) - f(a)}{x - a}\), the gap \(f(x) - L(x)\) shrinks faster than \(x - a\) itself as \(x \to a\). For twice-differentiable functions the error is roughly \(\tfrac{1}{2}f''(a)(x-a)^2\): halve the distance, quarter the error. That quadratic error term will reappear in Taylor polynomials and as the reason Newton’s method converges so quickly.

Take \(f(x) = \sqrt{x} = x^{1/2}\) and \(a = 4\). Try it: Estimate √4.1 with a tangent.

The true value is \(2.0248457\ldots\), so the error is about \(1.5\times 10^{-4}\). Check against the error estimate: \(f''(x) = -\tfrac{1}{4}x^{-3/2}\), so \(f''(4) = -\tfrac{1}{32}\), and \(\tfrac{1}{2}\cdot\tfrac{1}{32}\cdot(0.1)^2 \approx 1.56\times 10^{-4}\). The estimate also predicts the sign: since \(\sqrt{x}\) is concave down, the tangent lies above the curve and the approximation overshoots.

Linearise \(\sin x\) at \(a = 0\): \(f(0) = 0\), \(f'(0) = \cos 0 = 1\), so \(L(x) = x\). For small angles in radians, \(\sin x \approx x\). At \(x = 0.1\) (about \(5.7^\circ\)), \(\sin 0.1 = 0.0998334\), an error under \(0.2\%\). Try it: sin x ≈ x, visually.

This is the step every Mechanics course takes with the simple pendulum. The exact equation of motion contains \(\sin\theta\); replacing it with \(\theta\) turns it into the equation of simple harmonic motion, whose solutions are sines and cosines with period \(2\pi\sqrt{\ell/g}\). That famous formula is only true for small swings — it is a linear approximation.

You measure a ball’s radius as \(10\) cm, with a possible error of \(\pm 0.1\) cm. How uncertain is the computed volume?

Relative to \(V = \tfrac{4}{3}\pi(1000) \approx 4189\ \text{cm}^3\), that is

A 1% error in the radius becomes a 3% error in the volume: cubing triples relative errors. (In general, for \(y = x^n\), relative error multiplies by \(n\).)

In Chapter 16 you met limits like \(\lim_{x\to 0}\frac{\sin x}{x}\), where plugging in gives \(\frac{0}{0}\) — an indeterminate form, meaning the answer depends on how fast numerator and denominator approach zero. Algebraic tricks and squeezing work sometimes. Is there a general tool? Yes: compare their rates of approach — their derivatives.

Suppose \(f(a) = g(a) = 0\) and both are differentiable at \(a\) with \(g'(a) \ne 0\). Near \(a\), replace each by its linearisation:

Both functions start at zero; their ratio is governed by their speeds. The full theorem (proved with a generalisation of the Mean Value Theorem) states:

provided the right-hand limit exists (or is \(\pm\infty\)). It also works for \(a = \pm\infty\) and one-sided limits.

1. \(\displaystyle\lim_{x\to 0}\frac{\sin x}{x}\): form \(\frac00\).

(Honesty note: the derivative of \(\sin x\) was itself derived using this limit in Chapter 17, so this is a consistency check, not an independent proof.)

2. \(\displaystyle\lim_{x\to 0}\frac{e^x - 1 - x}{x^2}\): form \(\frac00\). Apply twice:

3. \(\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}\): form \(\frac{\infty}{\infty}\).

Logarithms grow more slowly than any line — the calculus behind "\(O(\log n)\) beats \(O(n)\)".

4. \(\displaystyle\lim_{x\to 0^+} x\ln x\): form \(0\cdot(-\infty)\). Rewrite as a quotient first:

This limit is why information theory can define \(0\log 0 = 0\) in the entropy formula.

A tangent line matches a function’s value and slope at one point. It is good nearby, and then drifts off. Could we do better by matching the bending too — the second derivative — and the third, and so on? And how do computers actually evaluate \(\sin x\), when the only operations a chip performs are addition and multiplication?

Polynomials are the functions computers can evaluate with just \(+\) and \(\times\). So look for a polynomial

that agrees with \(f\) at \(x = 0\) in value and in its first \(n\) derivatives. Differentiate \(p\) repeatedly and set \(x = 0\); each time, every term vanishes except one:

The \(k!\) (k factorial, \(k! = 1\cdot 2\cdots k\), from Chapter 24 and Chapter 14) appears because differentiating \(x^k\) \(k\) times brings down \(k, k-1, \ldots, 1\). The result is the Taylor polynomial of degree \(n\) about \(0\) (also called a Maclaurin polynomial):

About a general centre \(a\), replace \(x\) by \((x - a)\) and evaluate derivatives at \(a\). Degree 1 is exactly the linearisation \(L(x)\).

The derivatives of \(\sin x\) cycle with period 4: Try it: Add Taylor terms to sin x; Taylor series of sin x.

Substituting into the formula, the even terms vanish:

At \(x = 0.5\) (where \(\sin 0.5 = 0.4794255\ldots\)):

Each extra term gains about two decimal digits here. The same procedure gives the other two essential expansions:

(for \(e^x\), every derivative is \(e^x\), which is 1 at 0, so every coefficient is \(1/k!\)).

A short program shows the pattern — accuracy is excellent near the centre and poor far away:

Each new term is computed from the previous one (multiply by \(-x^2/((2k+2)(2k+3))\)), never from scratch — the factorials are never formed explicitly.

Try it: Where ln(1 + x) stops converging.

Many questions reduce to finding a root (or zero) of a function: a number \(r\) with \(f(r) = 0\).

Any equation \(\text{left} = \text{right}\) can be rewritten as \(\text{left} - \text{right} = 0\), so "find a root" is the universal form of "solve an equation".

For linear and quadratic equations we have formulas (Chapter 5, Chapter 8). Cubics and quartics have formulas too, but they are monstrous. For polynomials of degree five and higher, the Abel–Ruffini theorem says there is no general formula in radicals at all. And equations that mix kinds of function, like \(\cos x = x\) or \(e^x = 3x\), typically have no closed-form solution in terms of familiar functions.

Yet these equations have perfectly definite answers, and we need them to 15 digits. The way forward is to stop looking for a formula for the answer, and instead look for a procedure that improves a guess.

We cannot easily solve \(f(x) = 0\) for a curved \(f\). But we can trivially solve \(L(x) = 0\) for a straight line \(L\). And from Section 18.6 we know that near any point, a smooth curve is well approximated by its tangent line.

So: stand at your current guess \(x_n\) on the curve, at the point \((x_n, f(x_n))\). Draw the tangent there. Follow it down (or up) to where it crosses the \(x\)-axis. Because the tangent hugs the curve, the place where the tangent hits zero should be close to where the curve hits zero — closer than where you started. Take that crossing as the new guess and repeat.

The tangent at \(x_n\) has slope \(f'(x_n)\) and passes through \((x_n, f(x_n))\). In point-slope form:

Call its \(x\)-intercept \(x_{n+1}\). At that point \(y = 0\):

That is Newton’s method (also called the Newton–Raphson method):

Read it in words: the correction is the current height divided by the current slope. If you are at height \(f(x_n) = 6\) on a slope of \(6\), a straight line would reach zero after stepping back \(1\) unit. The minus sign points the step in the right direction automatically: positive height with positive slope means step left, and so on. Notice also the division by \(f'(x_n)\) in the third line — we have already spotted the method’s Achilles' heel.

Newton’s method needs a starting guess \(x_0\), and the quality of that guess matters more than for any other step. Good ways to get one:

Avoid starting near points where \(f'(x) = 0\) — turning points of the curve — for reasons we will see in step 10.

Now do it by hand. Find \(\sqrt{2}\) using \(f(x) = x^2 - 2\), whose derivative is \(f'(x) = 2x\), starting from \(x_0 = 1\) (since \(1^2 = 1 < 2 < 4 = 2^2\), the root is between 1 and 2).

Substituting \(f\) and \(f'\) into the general formula first gives a specific iteration:

Iteration 1 (\(n = 0\)):

Iteration 2 (\(n = 1\)):

Iteration 3 (\(n = 2\)). Working in fractions keeps it exact:

Iteration 4 (\(n = 3\)):

Compare with the true value \(\sqrt{2} = 1.414\,213\,562\,373\,095\ldots\):

Try it: Newton versus bisection for √2.

Look at the last column: 1, 3, 6, 12. The number of correct digits roughly doubles every step. Four steps of arithmetic you could do on paper give twelve correct digits.

An algorithm needs a termination condition. Since we don’t know the true root (if we did, we wouldn’t be computing it), we stop on evidence available inside the loop:

Do not ask for more accuracy than the arithmetic can deliver. Double-precision floats carry about 16 significant digits (Chapter 32); a tolerance of \(10^{-20}\) on a root near 1 can never be met.

Try it: Step Newton’s cubic. ==== Worked example: Newton’s own cubic

Solve \(x^3 - 2x - 5 = 0\). This is the very equation Newton used to demonstrate his method around 1669; it has no rational root and one real root.

Set up. \(f(x) = x^3 - 2x - 5\), \(f'(x) = 3x^2 - 2\).

Initial value. Try integers: \(f(2) = 8 - 4 - 5 = -1\) and \(f(3) = 27 - 6 - 5 = 16\). The sign change puts a root between 2 and 3, and \(|f(2)|\) is much smaller, so take \(x_0 = 2\).

Iteration 1:

Iteration 2:

Iteration 3:

Iteration 4: \(f(x_3) \approx 1.74\times 10^{-9}\), \(f'(x_3) \approx 11.1614\), correction \(\approx 1.56\times 10^{-10}\), giving \(x_4 = 2.094\,551\,481\,542\,3\).

Stop? The last step was \(1.6\times 10^{-10}\); with \(\varepsilon = 10^{-8}\) we stop. The root is \(x \approx 2.094\,551\,48\). The corrections shrank \(0.1 \to 5.4\times 10^{-3} \to 1.7\times 10^{-5} \to 1.6\times 10^{-10}\) — again, the exponent roughly doubles.

The procedure translates almost line for line. We record every iterate so that we can study convergence.

Three things to notice. The hand calculation and the program agree digit for digit. After \(x_5\) the error is at the level of floating-point rounding — \(x_6\) differs from \(x_5\) by one unit in the last place, and that tiny step is what triggers the stop. And the last call, which we will explain shortly, never converges at all: the iteration cap turned an infinite loop into a clear error message.

Why do the digits double? Let \(r\) be the root and write the error as \(e_n = x_n - r\). Expand \(f\) around \(x_n\) using a Taylor polynomial of degree 1 plus its error term (Section 18.8), evaluated at the root:

for some \(\xi\) between \(x_n\) and \(r\). Divide by \(f'(x_n)\) and rearrange:

The new error is proportional to the square of the old one. If the error is \(10^{-3}\), the next is about \(C\times 10^{-6}\), then \(C^3\times 10^{-12}\): the exponent doubles. This is quadratic convergence.

Check it on \(\sqrt2\): \(C \approx \frac{f''(r)}{2f'(r)} = \frac{2}{2\cdot 2\sqrt2} \approx 0.3536\). From \(e_3 = 2.124\times 10^{-6}\), the prediction is \(e_4 \approx 0.3536\times(2.124\times 10^{-6})^2 = 1.595\times 10^{-12}\), and the measured error was \(1.595\times 10^{-12}\).

The formula also explains what "works well" requires: \(f'(r) \ne 0\) (a simple root, so \(C\) is finite), \(f''\) not wild, and \(x_0\) close enough that \(|C e_0| < 1\). Under those conditions Newton’s method is the fastest practical root finder there is.

Try it: Race the two methods.

Quadratic convergence is a local promise: it holds once you are close enough. Far from a root, or at a badly behaved one, the tangent line is a poor model of the curve, and following it can lead anywhere. These are not curiosities; you will meet every one of them in practice.

Oscillation: \(f(x) = x^3 - 2x + 2\) from \(x_0 = 0\). Here \(f'(x) = 3x^2 - 2\). Try it: Watch the 0–1 cycle.

We are back where we started; the iteration cycles \(0, 1, 0, 1, \ldots\) forever. Worse, the cycle attracts nearby starts: from \(x_0 = 0.1\) the iterates are drawn into the same 0–1 cycle. The only real root is near \(-1.769\), on the other side of a hump. Start at \(x_0 = -2\) instead and Newton gives \(-1.8,\ -1.76995,\ -1.7692927,\ -1.769292354\) — converged in four steps.

Divergence: the iterates run away. For \(f(x) = \sqrt[3]{x} = x^{1/3}\) (root at 0), \(f'(x) = \tfrac13 x^{-2/3}\), and Try it: Find the arctan threshold; Newton diverging on ∛x.

Every step doubles the distance from the root, from any nonzero start: \(0.5, -1, 2, -4, \ldots\). The curve is so steep at its root (vertical tangent) that the tangents from elsewhere overshoot. A less contrived example: \(f(x) = \arctan x\) from \(x_0 = 1.5\) gives \(-1.694,\ 2.321,\ -5.114,\ 32.30,\ -1575.3, \ldots\) — the curve flattens out away from the root, so each tangent shoots further. (From \(x_0 = 1\) it converges; the boundary is near \(x_0 \approx 1.39\).)

Flat tangent: \(f'(x_n) \approx 0\). The correction \(f(x_n)/f'(x_n)\) divides by the slope; near a turning point the slope is tiny and the step is huge. For \(x^2 - 2\) from \(x_0 = 0.01\): Try it: The flat-tangent leap.

Here the parabola is kind and the iterates crawl back (\(50.01, 25.03, 12.55, 6.36, 3.34, 1.97, 1.49, 1.416, \ldots\)), costing 11 steps instead of 5. On a function with several roots the giant leap usually lands near a different root, or somewhere the method then diverges. Exactly at \(f'(x_n) = 0\) the tangent is horizontal, never meets the axis, and the formula divides by zero.

Wrong root, and sensitive dependence on \(x_0\). \(f(x) = x^3 - x = x(x-1)(x+1)\) has roots \(-1, 0, 1\). You might expect Newton to find the nearest one. Not so: starting from \(0.44\) reaches \(0\); from \(0.45\) it reaches \(-1\); from \(0.46\) it reaches \(+1\); from \(0.5\) it reaches \(-1\) again. Near \(x = \pm 1/\sqrt{3} \approx \pm 0.577\), where \(f' = 0\), tiny tangents throw the iterate across the graph. The set of starting values that lead to a given root is that root’s basin of attraction, and the basins interleave in ever finer bands near \(\pm 1/\sqrt5 \approx \pm 0.447\). Try it: Zoom into the basins.

Slow convergence at a repeated root. For \(f(x) = (x - 1)^2\), the root \(x = 1\) is a double root where \(f'(1) = 0\) too. The iteration becomes \(x_{n+1} = x_n - \frac{(x_n - 1)^2}{2(x_n - 1)} = x_n - \frac{x_n - 1}{2}\): the error merely halves each step. Newton degrades to linear convergence — no better than bisection — whenever the curve touches the axis instead of crossing it. Try it: Newton at a double root.

Newton’s method is one of several ways to solve \(f(x) = 0\). The most important companion is bisection.

Bisection. If \(f(a)\) and \(f(b)\) have opposite signs, a root lies in \(\left[a, b\right]\). Evaluate the midpoint \(m = \frac{a+b}{2}\); whichever half still has a sign change contains a root; keep that half and repeat.

Forty steps for twelve digits, against Newton’s four or five. But bisection cannot fail once you have a bracket: the interval halves every time, so after \(n\) steps the error is at most \((b - a)/2^{n}\). It is binary search (Chapter 12) applied to a continuous function.

Secant method. Replace \(f'(x_n)\) by the slope of the secant through the last two iterates: \(x_{n+1} = x_n - f(x_n)\dfrac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})}\). No derivative needed; order \(\approx 1.618\); same failure modes as Newton.

Closed-form algebra. If a formula exists, use it. \(x^2 - 5x + 6 = 0\) practically begs to be factored as \((x - 2)(x - 3) = 0\); running Newton’s method on it would be rather silly.

Choosing: use algebra when it applies. When it doesn’t: if you have a good starting guess and a cheap derivative (or need to solve many similar equations — e.g. a square root for every pixel), use Newton. If you know nothing about where the root is, or reliability matters more than speed, bracket and bisect. In production code, use a hybrid: library routines such as Brent’s method take a fast Newton- or secant-like step when it stays inside the bracket and a bisection step when it doesn’t, getting both speed and a guarantee. Understanding the two pure methods is how you understand — and debug — the hybrid.