Before the final

Why is a tangent line such a good estimate near its point?

Estimate a cube root from a tangent line, then magnify the gap between line and curve: it starts flat, grows like the square of the step, and sits on the side the curve bends away from.

Builds on the definition of the derivative and the chain rule and logarithms.

Base point, a: 8
Estimate, L(30): 3.8333
True value, ∛30: 3.1072
Gap, L − f: 0.726

Step 1 of 4: Build it where f is easy

To estimate $\sqrt[3]{30}$ by hand, use the tangent line of $f(x) = \sqrt[3]{x}$ at a base point $a$ : $L(x) = f(a) + f'(a)(x - a)$ . It needs $f(a)$ and $f'(a)$ . From $a = 8$ , where $f(8) = 2$ , it gives $3.833$ , off by $0.73$ : the gap, drawn below.

Predict first. Which base point gives the best estimate you can work out by hand?

1. Build it where f is easy
To estimate $\sqrt[3]{30}$ by hand, use the tangent line of $f(x) = \sqrt[3]{x}$ at a base point $a$ : $L(x) = f(a) + f'(a)(x - a)$ . It needs $f(a)$ and $f'(a)$ . From $a = 8$ , where $f(8) = 2$ , it gives $3.833$ , off by $0.73$ : the gap, drawn below.
Question: Which base point gives the best estimate you can work out by hand?
$a = 27$ . It is a perfect cube, so $f(27) = 3$ and $f'(27) = \frac{1}{3 \cdot 27^{2/3}} = \frac{1}{27}$ come by hand, and it is close to $30$ : $L(30) = 3 + \frac{3}{27} = 3.111$ . A line built at $30$ would need $\sqrt[3]{30}$ itself, the number being estimated.
The figure: The curve f(x) = ∛x and its tangent line at the base point a = 27, where f(27) = 3. At x = 30 the line gives 3.1111 and the curve 3.1072. The gap between the tangent line at 27 and the curve, L(x) − f(x): 0 at x = 27 and above 0 on both sides. At x = 30 it is 0.00388.
2. The step from the base point
A common slip multiplies the slope by $x$ rather than by the step from the base point, and writes $\sqrt[3]{30} \approx 3 + \frac{1}{27} \cdot 30 = 4.111$ . Its line, in amber, has the tangent's slope but misses the curve altogether.
Question: Which number should the slope $\frac{1}{27}$ multiply?
$3$ , the step $x - a$ . The slope is the rise of $f$ per unit of run from the base point, so the run is measured from $27$ . Measured from $0$ instead, the line moves up by $\frac{1}{27} \cdot 27 = 1$ : the amber line is $1$ too high at every $x$ , and gives $4.111$ for a number near $3.1$ .
The figure: The curve f(x) = ∛x and its tangent line at the base point a = 27, where f(27) = 3. At x = 30 the line gives 3.1111 and the curve 3.1072. In amber, the slip's line, parallel to the tangent and 1 higher, which gives 4.1111. The gap between the tangent line at 27 and the curve, L(x) − f(x): 0 at x = 27 and above 0 on both sides. At x = 30 it is 0.00388. In amber, the slip's gap, about 1 everywhere.
3. Too high or too low
Close up, the line and the curve are hard to tell apart; the lower panel magnifies the gap between them. $L(30) = 3.111$ is too high, as $\sqrt[3]{30} = 3.107$ .
Question: Why is the estimate too high?
It bends down: $f''(x) = -\frac{2}{9}x^{-5/3}$ is negative, so the curve falls away below every tangent line, on both sides. Step down from $27$ to $24$ : $L(24) = 2.889$ and $\sqrt[3]{24} = 2.884$ , too high again. Where $f'' > 0$ a curve bends up, and tangent estimates come out too low.
The figure: The curve f(x) = ∛x and its tangent line at the base point a = 27, where f(27) = 3. At x = 24 the line gives 2.8889 and the curve 2.8845. The gap between the tangent line at 27 and the curve, L(x) − f(x): 0 at x = 27 and above 0 on both sides. At x = 24 it is 0.00439.
4. The gap grows like the square
The gap leaves the base point flat and grows slowly at first. At $30$ , a step of $3$ from $27$ , it is $0.0039$ . How far must the step go before the gap is four times as big?
Drag the target to the right until the gap is four times its size at $30$ .
A step of about $6$ : twice the step, four times the gap. At $33$ the gap is $0.0147$ . It grows like the square of the step, about $\frac12|f''(27)|(x - 27)^2$ , which is why a tangent line is so good close to its point, and why the base point should be near the value you want.
$f(x) \approx f(a) + f'(a)(x - a), \quad \text{off by about } \tfrac12\,|f''(a)|\,(x - a)^2$
The figure: The curve f(x) = ∛x and its tangent line at the base point a = 27, where f(27) = 3. At x = 33.2 the line gives 3.2296 and the curve 3.2140. The gap between the tangent line at 27 and the curve, L(x) − f(x): 0 at x = 27 and above 0 on both sides. At x = 33.2 it is 0.0156.

Step 1 of 4: Build it where f is easy

The lesson as text

1. Build it where f is easy

2. The step from the base point

3. Too high or too low

4. The gap grows like the square