Linear approximation: f(a+h) ≈ f(a) + h·f'(a)

The key formula

If $f$ is differentiable at $a$, then for small $h$:

$f(a + h) \approx f(a) + h \cdot f'(a)$

That’s the equation of the tangent to the graph at $a$, used locally to approximate $f$.

Geometric interpretation

Near the point $A(a, f(a))$, the curve and tangent line are nearly identical. The tangent gives a good estimate of $f(a + h)$.

The smaller $h$, the better the approximation.

Quick computation examples

$\sqrt{4.01}$? Take $f(x) = \sqrt{x}$, $a = 4$, $h = 0.01$.

$f(4) = 2$, $f'(x) = \dfrac{1}{2\sqrt{x}}$, $f'(4) = 1/4$.

$\sqrt{4.01} \approx 2 + 0.01 \times 1/4 = 2 + 0.0025 = 2.0025$.

Actual value: $2.002498\ldots$ ✓

$(1.02)^{10}$? $f(x) = x^{10}$, $a = 1$, $h = 0.02$.

$f(1) = 1$, $f'(1) = 10$, so $(1.02)^{10} \approx 1 + 0.2 = 1.2$ (actual $\approx 1.219$).

Equivalent form with $x$ near $a$

$f(x) \approx f(a) + (x - a) \cdot f'(a)$

Taylor expansion (calculus)

Linear approximation is a 1st-order Taylor expansion. You can go further:

$f(a + h) \approx f(a) + h f'(a) + \dfrac{h^2}{2} f''(a) + o(h^2)$

Differential notation

$df \approx f'(a) \, dh$

Physical: small change in $f$ ≈ derivative × small change in variable.

Common mistakes

1. Using the formula with $h$ too large (poor approximation).
2. Forgetting the $f'(a)$ factor (just $f(a) + h$ is wrong).
3. Confusing $h$ and $x$ in the formula.

?À toi de jouer

ln(1.01) ≈ ? (f(x) = ln(x), a = 1)

A.0.01B.0.1C.1.01D.0

Valider