#### To determine

**To estimate:** The value f(0.9) and f(0.1).

#### Answer

The values f(0.9)≈4.8 and f(1.1)≈5.2.

#### Explanation

**Given:**

The value f(1)=5.

**Result used:**

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

**Calculation:**

Obtain the value of f(0.9).

The linearization of the function f(x) at x=0 is computed as follows,

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

f(x)≈f(1)+f′(1)(x−1)

Form the given graph, it is observed that f′(1)=2.

Substitute f′(1)=2 and f(1)=5,

f(x)≈5+2(x−1)=5+2x−2=2x+3

Therefore, the function f(x)≈2x+3 is true whenever the value of *x* is near to 1.

Substitute x=0.9 in f(x)≈2x+3,

f(0.9)≈2(0.9)+3=1.8+3=4.8

Thus, the value of f(0.9)≈4.8.

Substitute x=1.1 in f(x)≈2x+3,

f(1.1)≈2(1.1)+3=2.2+3=5.2

Thus, the value of f(1.1)≈5.2.

Therefore, the values f(0.9)≈4.8 and f(1.1)≈5.2.

#### To determine

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

#### Explanation

From the graph, it is observed that the value of derivative of the function is positive and the derivative of the function is decreasing.

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

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

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