#### To determine

**To explain:** The linear approximation is reasonable.

#### Explanation

**Result used:**

The linear approximation of the function at x=a is L(x)=f(a)+f′(a)(x−a).

**Calculation:**

Obtain the value of 4.02 by using linearization.

Since 4 is near integer of the value 4.02, choose the value a=4 and the function f(x)=x.

The linearization of the function f(x)=x at a=4 is computed as follows,

Consider f(x)=x,

Differentiate with respect to *x*,

f′(x)=ddx(x)=12x12−1=12x−12=12x

Substitute x=4,

f′(0)=124=14

Substitute x=4 in f(x)=x, f(4)=2.

Substitute the value a=4 in L(x)=f(4)+f′(4)(x−4),

L(x)=f(4)+f′(4)(x−4)=f(4)+f′(4)(x−4)

Substitute f′(4)=14 and f(4)=2,

L(x)=2+14(x−4)=2+14x−1=14x+1

Thus, the linearization of function at a=4 is L(x)=14x+1.

That is, x≈14x+1 when the value of *x* is near to 4.

Substitute x=4.02 in x≈14x+1,

4.02≈14(4.02)+1=14(4.02)+1=1.005+1=2.005

Hence, the approximation value is reasonable.