In this section, we introduce linear approximation of a function at a point.

**Linear
Approximation and Differentials**

**Linear
Approximation**

In this
section, we introduce linear approximation of a function at a point. Using the
linear approximation, we shall estimate the function value near a chosen point.
Then we shall introduce differential of a real-valued function of one variable,
which is also useful in applications.

Let *f* : (*a*,
*b*) → ℝ be a differentiable function and *x* ∈(*a*,
*b*) . Since *f *is differentiable at *x*
, we* *have

where ≈
means “**approximately”**
equal. Also, observe that as the independent variable changes from *x *to*
x *+* x*Δ, the function value changes from* f *(* x*)* *to*
f *(* x *+* *Δ*x*)* *. Hence if* *Δ*x *is small and the* *change in the output is denoted by Δ*f* or Δ*y* , then (2) can be rewritten as

change
in the output =* *Δ*y *=* *Δ*f *=* f *(* x *+ Δx) − *f (x)* ≈ *f ‘(x) *Δ*x*.

Note
that (3) helps in approximating the value of *f *(* x *+* *Δ*x*)* *using* f *(* x*)* *and*
f* *‘* ( *x*) Δ*x* . Also, for a fixed
*x*_{0} , *y*( *x* ) =
*f* ( *x*_{0} ) + *f*
′(
*x*_{0} )( *x* − *x*_{0} ), *x* ∈ **R**, gives the tangent line for the graph of *f *at* *(* x*_{0}* *,* f *(* x*_{0}* *)) which gives a good approximation to the function *f* near *x*_{0} . This leads us to define

**Definition 8.1 (Linear
Approximation)**

Let *f *: (*a*,*
b*)* *→* **ℝ** *be a differentiable function and* x*_{0}* *∈(*a*,* b*)*
*. We define the linear approximation *L
*of *f *at* x*_{0}* *by

*L *(* x*)* *=* f *(* x*_{0}* *)* *+* f *′(* x*_{0}* *)(*
x *−* x*_{0}* *),* *∀*x *∈* *(*a*,* b*) ... (4)

Note
that by (3) and (4) we see that

*f *(* x *+* *Δ*x*) ≈ *f (x)* +* f *′ (*x*)* *Δ*x
*,

which is
useful in approximating the value of *f *(* x *+* *Δ*x*).

Note
that linear approximation for *f* at *x*_{0}* *gives a good approximation to*
f *(* x*)* *if* x *is close to *x*_{0} , because

Tangent
line *x* =* f *(x_{0}) +* f *′(x_{0})(*x*−*x*_{0})

Error = *f* (*x*)
−
*L* ( *x*) = *f*
(*x* ) − *f*
(*x*_{0} ) −
*f* ′(*x*_{0}
)(*x* − *x*_{0}
) ... (5)

approaches
zero as *x* approaches to *x*_{0} by continuity of *f* at *x*_{0}
. Also, if *f* ( *x*) = *mx*
+
*c* , then its linear approximation is *L* ( *x*)
=
(*mx*_{0} +
*c* ) + *m*(*x* − *x*_{0}
) =
*mx* + *c*
=
*f* (*x*) , for any point *x* ∈(*a*,
*b*) . That is, the linear
approximation, in this case, is the original function itself (is it not
surprising?).

** **

**Example 8.1**

Find the
linear approximation for *f (x)* = √(1+*x*)*
*,* x *≥ −1, at *x*_{0} = 3. Use the linear approximation to estimate *f* (3.2) .

**Solution**

We know
from (4), that L(*x*) =* f *(*x*_{0})
+* f *′(*x*_{0})(*x* − *x*_{0}) . We have *x*_{0} = 3,* *Δ*x *= 0.2 and hence *f* (3) = √(1+ 3) = 2 . Also,

Actually,
if we use a calculator to calculate we get √4.2 = 2.04939.

Tags : Mathematics , 12th Maths : UNIT 8 : Differentials and Partial Derivatives

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