#### To determine

**To estimate:** The value g(1.95) and g(2.05).

#### Answer

The values g(1.95)=−4.15 and g(2.05)=−3.85.

#### Explanation

**Given:**

The value g(2)=−4 and g′(x)=x2+5.

**Result used:**

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

**Calculation:**

The linearization of the function g(x) at x=2 is computed as follows,

Substitute the value a=2 in g(x)≈g(a)+g′(a)(x−a),

g(x)≈g(2)+g′(2)(x−2)

Given g′(x)=x2+5 is,

Substitute x=2,

g′(2)=22+5=4+5=9=3

Substitute g(2)=−4 and g′(2)=3 in g(x)≈g(2)+g′(2)(x−2),

g(x)≈−4+3(x−2)=−4+3x−6=3x−10

Therefore, the function g(x)=3x−10 is true whenever the value of *x* is near to 2.

Substitute x=1.95 in g(x)=3x−10,

g(1.95)=3(1.95)−10=5.85−10=−4.15

Thus, the value of g(1.95)=−4.15.

Substitute x=2.05 in g(x)=3x−10,

g(2.05)=3(2.05)−10=6.15−10=−3.85

Thus, the value of g(2.05)=−3.85.

Therefore, the values g(1.95)=−4.15 and g(2.05)=−3.85.

#### To determine

**To explain:** Whether the estimate value in part (a) is too large or too small.

#### Answer

The estimated value in part (a) is too small.

#### Explanation

The graph of the derivative g′(x) as shown below in Figure 1.

From the graph, it is observed that the value of derivative of the function is positive and the derivative of the function is increasing near to value of 2.

That is, the slope of the tangent line is positive and it becomes more steep.

Thus, the tangent line lies on the below the curve.

Therefore, the estimated value in part (a) is too small.