Concept preview

What you will be able to do

1. 01State and use the linearization $L(x)=f(a)+f'(a)(x-a)$ — the tangent line at $x=a$ — to approximate values of a differentiable function near the base point $a$.
2. 02Choose a convenient base point $a$ where $f(a)$ and $f'(a)$ are easy, and estimate quantities such as $\sqrt{4.2}$ or $\sqrt[3]{8.1}$, then decide from the concavity whether the estimate is an over- or under-estimate.
3. 03Connect linearization to differentials $dy=f'(x)\,dx$ and use them to approximate a small change $\Delta y$ in an output.

Suppose $f$ is differentiable at $x=a$. The tangent line there passes through the point $\big(a,\,f(a)\big)$ with slope $f'(a)$, so by point-slope form its equation is $y-f(a)=f'(a)(x-a)$. Writing this as a function of $x$ gives the LINEARIZATION of $f$ at $a$: $L(x)=f(a)+f'(a)(x-a).$ Read the formula as instructions: START at the known height $f(a)$, then WALK along the slope $f'(a)$ for a run of $(x-a)$. Because the curve hugs its tangent near $a$, we have the approximation $f(x)\approx L(x)$ whenever $x$ is close to $a$.