#### To determine

**To find:** The linear approximation of the function f(x)=1−x at a=0 and use it to approximate the numbers 0.953 and 1.13.

#### Answer

The linear approximation of the function at x=0 is g(x)≈1+13x and the numbers are 0.953≈−0.983¯ and 1.13≈1.03¯.

#### Explanation

**Given:**

The function is g(x)=1+x3 and the point a=0.

**Result used:**

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

**Derivative rules:** Chain rule

If y=f(u) and u=g(x) are both differentiable function, then dydx=dydu⋅dudx.

**Calculation:**

Consider g(x)=1+x3.

Differentiate with respect to *x*,

g′(x)=ddx(1+x3)=ddx((1+x)13)

Substitute u=1+x,

g′(x)=ddx((u)13)

Apply the chain rule and simplify the terms,

g′(x)=ddu((u)13)dudx=13u13−1dudx=13u−23dudx

Substitute u=1+x in g′(x),

g′(x)=13(1+x)−23ddx(1+x)=13(1+x)−23[ddx(1)+ddx(x)]=13(1+x)−23[0+1]=13(1+x)−23

Thus, the derivative is g′(x)=13(1+x)−23.

Substitute x=0 in g′(x)=13(1+x)−23,

g′(0)=13(1+0)−23=13

Thus, the value is g′(0)=13.

Substitute x=0 in g(x)=1+x3,

g(0)=1+03=1

Thus, the value is g(0)=1.

Substitute the value a=0 in g(x)≈g(a)+g′(a)(x−a),

g(x)≈g(0)+g′(0)(x−0)=g(0)+xg′(0)

Substitute g′(0)=13 and g(0)=1,

g(x)≈1+13x

Therefore, the linear approximation of the function at x=0 is g(x)≈1+13x.

Use linear approximation to compute the number 0.953,

Since g(x)≈1+13x, that is 1+x3≈1+13x.

0.953=1−0.053

Here, the value x=0.05 since g(x)≈1+13x is true when the value of *x* is near to 0.

0.953≈1+13(−0.05)=1−0.16¯=−0.983¯

Therefore, the value of 0.953 is approximately −0.983¯.

1.13=1+0.13

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

1+0.13≈1+13(0.1)=1+0.03¯=1.03¯

Therefore, the value of 1.13 is approximately 1.03¯.

**Graph:**

Using online graphic 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 g(x)=1+x3 is g(x)≈1+13x.