Concept understanding — Linear Approximation and Differentials
A nonlinear function is generally hard to evaluate exactly, but near any one point its graph looks almost like a straight line -- the tangent line at that point. This is the idea behind linear approximation.
Definition (Linear Approximation). Let f:(a,b)→R be differentiable and x0∈(a,b). The linear approximation L of f at x0 is
L(x)=f(x0)+f′(x0)(x−x0),∀x∈(a,b).
This is exactly the equation of the tangent line to y=f(x) at (x0,f(x0)). Because f is differentiable, f(x0+Δx)≈L(x0+Δx)=f(x0)+f′(x0)Δx when Δx is small, and the errorf(x)−L(x) shrinks to 0 faster than x→x0 (it is o(x−x0), a consequence of the definition of the derivative as a limit).
Tip
To linearly approximate an "ugly" value like 327.2 or (123)2/3: pick a nearby point x0 where the function and its derivative are easy to evaluate exactly (a perfect cube, a perfect fourth power, a multiple of 10, ...), compute f(x0) and f′(x0), and plug into L(x0+Δx)=f(x0)+f′(x0)Δx.
The differential. Writing Δx=dx, the tangent-line increment is called the differential of f:
df=f′(x)dx,equivalently df=f′(x)Δx.
Geometrically, Δf=f(x+dx)−f(x) is the actual rise along the curve, while df=f′(x)dx is the rise along the tangent line; for small dx, Δf≈df, but the two are generally not equal (only exactly equal when f is itself linear, f(x)=mx+c). df is a function of two independent quantities, x and dx -- not of x alone the way the derivative is.
Differentials of standard functions follow directly from the derivative rules, e.g. d(xn)=nxn−1dx, d(sinx)=cosxdx, d(ex)=exdx, d(logx)=x1dx. The algebraic properties of differentials mirror differentiation exactly:
and the differential is dF=∂x∂Fdx+∂y∂Fdy, with dx=Δx,dy=Δy. Geometrically this is the tangent plane to z=F(x,y) at (x0,y0) -- exactly as the one-variable linear approximation was a tangent line. The same pattern extends to three variables:
A linear approximation problem always has three ingredients: the base point (where the function is easy), the function value there, and the derivative(s) there. Get all three right and the rest is substitution.
Compute f(x0),f′(x0) and substitute into L(x)=f(x0)+f′(x0)(x−x0).
(i) f′(x)=3x2−5: f(2)=10,f′(2)=7⇒L=7x−4.
(ii) g′(x)=x/x2+9: g(−4)=5,g′(−4)=−54⇒L=−54x+59.
(iii) h′(x)=1/(x+1)2: h(1)=21,h′(1)=41⇒L=4x+1.
✓Final answer
(i) L(x)=7x−4 (ii) L(x)=−54x+59 (iii) L(x)=41x+41
Each part needs f(x0) and f′(x0) at the given point, then direct substitution into L(x)=f(x0)+f′(x0)(x−x0).
Part (i): f(x)=x3−5x+12,x0=2.
f(2)=8−10+12=10. f′(x)=3x2−5, so f′(2)=12−5=7.
L(x)=10+7(x−2)=10+7x−14=7x−4.
Part (ii): g(x)=x2+9,x0=−4.
g(−4)=16+9=25=5. g′(x)=2x2+92x=x2+9x, so g′(−4)=5−4.