4.6 Approximating Values of a Function Using Local Linearity and Linearization

4.6 Approximating Values of a Function Using Local Linearity and Linearization

Close to a point where you know a function's value and its slope, the function behaves almost like its tangent line. That makes the tangent line a practical tool: it turns a hard value such as $\sqrt{6}$ into simple arithmetic, and the shape of the graph tells you whether the estimate is too high or too low.

Local linearity: a curve becomes almost straight nearby

Near a point $x=a$, a smooth curve resembles its tangent line. The tangent line is therefore a locally linear approximation: a line that models the function accurately for inputs close to $a$. This is CHA-3.F.1 and supports Suggested Skill 1.F: Approximate values of a function using the equation of a tangent line.

If $f(a)$ and $f'(a)$ are known, the tangent line at $x=a$ is

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

The approximation is

$$ f(x)\approx L(x) $$

when $x$ is near $a$. The expression $L(x)$ is called the linearization of $f$ at $x=a$.

Worked example: estimating $\sqrt{6}$

Choose the nearby function $f(x)=\sqrt{x}$ and the convenient known point $a=4$, because $f(4)=2$. Since

$$ f'(x)=\frac{1}{2\sqrt{x}}, $$

we have

$$ f'(4)=\frac14. $$

Thus the linearization is

$$ L(x)=2+\frac14(x-4). $$

At $x=6$,

$$ \sqrt{6}\approx L(6) =2+\frac14(2) =2.5. $$

The actual value is approximately $2.449$, so this tangent-line estimate is an overestimate. That comparison is not an accident: $f(x)=\sqrt{x}$ is concave down near $x=4$, and a tangent line lies above a concave-down curve. In general, CHA-3.F.2 connects nearby concavity to approximation error: concave-up behavior produces a tangent-line underestimate, while concave-down behavior produces a tangent-line overestimate.

Misconception check — “A tangent line gives the exact value.”
The tangent line agrees exactly with the function at the point of tangency, $x=a$. Away from $a$, it gives an approximation whose accuracy depends on how close $x$ is to $a$ and how much the graph bends.

Retrieval check

A sphere has radius $r(t)$ and volume $V(t)=\frac43\pi r^3$. If $\frac{dr}{dt}$ is known, differentiate with respect to $t$ before inserting the current radius. For linearization, explain why a concave-up function produces a tangent-line underestimate: the graph bends upward and lies above its tangent line near the point of tangency.