Linear Approximation

Definition 8.1 (Linear Approximation). Let $f:(a,b)\to\mathbb R$ be differentiable and $x_0\in(a,b)$. The linear approximation $L$ of $f$ at $x_0$ is

$L$ is precisely the tangent line to $y=f(x)$ at the point $(x_0,f(x_0))$: it passes through that point (since $L(x_0)=f(x_0)$) with slope $f'(x_0)$. Because $f$ is differentiable at $x_0$, equation (1) above says $f(x_0+\Delta x)-f(x_0)\approx f'(x_0)\Delta x$, i.e.

which is exactly why $L$ is a good stand-in for $f$ near $x_0$. Writing the error of the approximation as

continuity of $f$ at $x_0$ forces this error $\to0$ as $x\to x_0$ — and in fact (from differentiability) the error goes to $0$ faster than $x-x_0$ does, which is the precise sense in which $L$ "hugs" $f$ near $x_0$.

A sanity check. If $f$ is itself linear, $f(x)=mx+c$, then $L(x)=mx_0+c+m(x-x_0)=mx+c=f(x)$ — the linear approximation of a linear function is the function itself, exactly as it should be.

Worked pattern. To find the linear approximation of $f$ at $x_0$ and use it to estimate $f(x_0+\Delta x)$: compute $f(x_0)$ and $f'(x_0)$ exactly, substitute into (4) to get $L(x)$, then evaluate $L$ at the required point. For example, for $f(x)=\sqrt{x+1}$ at $x_0=3$: $f(3)=2$, $f'(x)=\dfrac{1}{2\sqrt{x+1}}$ so $f'(3)=\dfrac14$; hence $L(x)=2+\tfrac14(x-3)=\tfrac{x}{4}+\tfrac54$, and $f(3.2)\approx L(3.2)=\tfrac{3.2}{4}+\tfrac54=0.8+1.25=2.05$ (the calculator value is $\sqrt{4.2}\approx2.04939$ — a very close match for a change of only $0.2$ in $x$). …