Close enough, and you can say which way AB & BC

If $f$ is differentiable at $a$, then near $x = a$ the graph of $f$ is nearly indistinguishable from its tangent line there. The tangent line, written as a function, is the linearization of $f$ at $a$:

$$L(x) = f(a) + f'(a)(x - a).$$

Three ingredients, and each is doing a job:

1. $f(a)$ is where the line starts, the height of the curve at the center.
2. $f'(a)$ is the slope, the rate at which the line climbs or falls.
3. $(x - a)$ is how far you moved from the center, and it is the piece that disappears most often.

Dropping it gives $L(x) = f(a) + f'(a)x$, a line with the right slope in the wrong place. To see how wrong, take $f(2) = 7$ and $f'(2) = 3$: the correct approximation of $f(2.1)$ is $7 + 3(0.1) = 7.3$, while the damaged version gives $7 + 3(2.1) = 13.3$, which is not close to anything.

Then $f(x) \approx L(x)$ for $x$ near $a$, and the approximation degrades as $x$ moves away.

The center $a$ has to satisfy two conditions at once: it must be close to the value being approximated, and $f(a)$ and $f'(a)$ must be easy to compute exactly. That is what makes the whole exercise worth doing.

1. To approximate $\sqrt{27}$, center at $a = 25$. Then $f(25) = 5$ and $f'(25) = \frac{1}{2\sqrt{25}} = 0.1$, both exact.
2. To approximate $\ln(1.05)$, center at $a = 1$, where $\ln 1 = 0$ and $\frac{1}{1} = 1$.
3. To approximate $\sqrt[3]{8.2}$, center at $a = 8$, where the cube root is exactly 2.

Two ways the choice goes wrong. Centering at the value being approximated is circular: to linearize $\sqrt{x}$ at $a = 27$ you would already need $\sqrt{27}$, which was the question. And centering somewhere merely convenient but distant, such as $a = e$ for $\ln(1.05)$, gives a legitimate line and a bad number, because the curve has drifted far from it.

So $\sqrt{27} \approx 5 + 0.1(27 - 25) = 5.2$, against a true value of $5.19615$. The error is under four thousandths on a two-unit step.

The tangent line lies on one side of the curve near the point of tangency, and which side is decided entirely by the second derivative.

1. $f'' < 0$ (concave down): the curve bends away from the line downward, so the line is above the curve. The approximation is an overestimate.
2. $f'' > 0$ (concave up): the curve bends away upward, so the line is below the curve. The approximation is an underestimate.

A concave-up curve is cupped like a bowl, and a straight line laid across a bowl passes underneath the rim on both sides of where it touches. That is the whole argument, and it is the same reasoning as Topic 2.3, where a forward-difference estimate of a derivative comes out high or low depending on which way the curve bends.

Two checks that both fail:

1. Using the sign of $f'$. Whether the function is rising or falling says nothing about which side of the curve the tangent is on. Both an increasing and a decreasing function can be concave up.
2. Assuming one answer always. "The tangent line always underestimates" is true for concave-up functions and false for every concave-down one, and $\sqrt{x}$ is concave down.

So $\sqrt{27} \approx 5.2$ is an overestimate, since $\sqrt{x}$ is concave down, and indeed $5.2 > 5.19615$. And $e^{0.1} \approx 1 + 0.1 = 1.1$ is an underestimate, since $e^{x}$ is concave up, and indeed $1.1 < 1.10517$.

One more setup error, easy to fall into once a table of values is on the page: using a secant slope where the tangent slope belongs.

The linearization is built from $f'(a)$, the instantaneous rate at the center. An average rate over a nearby interval is a different number and produces a different line. For $f(x) = x^{2}$ at $a = 3$, the tangent slope is $f'(3) = 6$ and the approximation of $f(3.1)$ is $9 + 6(0.1) = 9.6$, against a true $9.61$. Using the secant slope from $x = 2$ to $x = 3$, which is 5, gives $9.5$, and the error is ten times larger.

A checklist for every linearization, in order:

1. Is the center nearby and exactly computable?
2. Is the slope $f'(a)$, evaluated at the center, and not $f(a)$ by accident?
3. Is the $(x - a)$ present, and is it $x$ minus $a$ rather than the other way around?
4. Does the sign of $f''$ near $a$ say over or under?