#### 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 sec0.08 by using the linearization.

Since 0 is near integer of the value 0.08, choose the value a=0 and the function f(x)=secx.

Consider f(x)=secx,

Differentiate with respect to *x*,

f′(x)=ddx(secx)=secxtanx

Substitute x=0,

f′(0)=sec0tan0=1(0)=0

Substitute x=0 in f(x)=secx, f(0)=1.

Substitute the value a=0 in L(x)=f(a)+f′(a)(x−a),

L(x)=f(0)+f′(0)(x−0)=f(0)+f′(0)(x)

Substitute f′(0)=0 and f(0)=1,

L(x)=1+0(x)=1

Thus, the linearization of function at a=0 is L(x)=1.

That is, secx≈1 when the value of *x* is near to 0.08.

Substitute x=0.08 in secx≈1,

sec0.08≈1

Hence, the required result is reasonable.