#### To determine

**To find:** The linear approximation of the function f(x)=1−x at a=0 and use it to approximate the numbers 0.9 and 0.99.

#### Answer

The linear approximation of the function at a=0 is f(x)≈1−12x and the numbers are 0.9≈0.95 and 0.99≈0.995

#### Explanation

**Given:**

The function is f(x)=1−x and the point x=0.

**Result used:**

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

**Derivative rules:**

(1) *Chain rule:* If y=f(u) and u=g(x) are both differentiable function, then

dydx=dydu⋅dudx.

(2) *Quotient rule:* If f1(x) and f2(x) are both differentiable, then

ddx(f1(x)f2(x))=f2(x)ddx(f1(x))−f1(x)ddx(f2(x))[f2(x)].

**Calculation:**

Differentiate f(x)=1−x with respect to *x*,

f′(x)=ddx(1−x)

Let u=1−x and apply the chain rule,

f′(x)=ddu(u)dudx=12ududx

Substitute u=1−x,

f′(x)=121−xddx(1−x)=121−x(0−1)=−121−x

Substitute x=0 in f′(x) and obtain the value f′(0).

f′(0)=−121−0=−12

Thus, the value of f′(0)=−12.

Substitute x=0 in f(x) and compute the value of f(0).

f(0)=1−0=1

Thus, the value of f(0)=1.

Substitute the value a=0,f′(0)=−12 and f(0)=1 in f(x)≈f(a)+f′(a)(x−a),

f(x)≈f(0)+f′(0)(x−0)=1−12x

Therefore, the linear approximation of the function at a=0 is f(x)≈1−12x.

Use linear approximation to compute the number 0.9,

Since f(x)≈1−12x. That is, 1−x≈1−12x.

0.9=1−0.1

Here, the value of the *x* is 0.1.

0.9≈1−12(0.1)=1−0.5=0.95

Therefore the value of 0.9 is approximately 0.95.

Use linear approximation to compute the number 0.99,

0.99=1−0.01

Here, the value of the *x* is 0.01.

0.99≈1−12(0.01)=1−0.05=0.995

Therefore, the value of 0.99 is approximately 0.95.

**Graph:**

Using online graphing calculator to draw the curve and linear approximation of the function as shown below.

Form Figure 1, it is observed that the linear approximation of the function given f(x)=1−x is f(x)≈1−12x.