#### To determine

**To verify:** The linear approximation at
a=0 and to determine the value of *x* for which the linear approximation is accurate to within 0.1.

#### Answer

The values of *x* for which the linear approximation is accurate to within 0.1 is
−0.368<x<0.677.

#### Explanation

**Given:**

The function is
1+2x4≈1+12x and the point
a=0.

**Result used:**

The linear approximation of the function
g(x) 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 the function
f(x)=1+2x4.

Differentiate with respect to *x*,

f′(x)=ddx(1+2x4)

Let
u=1+2x,
f(x)=u14.

Apply the chain rule and simplify the terms,

f′(x)=ddu(u14)dudx=14u14−1dudx=14u−34dudx

Substitute
u=1+2x in
f′(x),

f′(x)=14(1+2x)−34d(1+2x)dx=14(1+2x)−34[d(1)dx+d(2x)dx]=14(1+2x)−34[0+2]=12(1+2x)−34

Thus, the derivative of the function is
f′(x)=12(1+2x)−34.

Substitute
x=0 in
f′(x)=12(1+2x)−34,

f′(0)=12(1+2(0))−34=12(1)=12

Thus, the value is
f′(0)=12.

Substitute
x=0 in
f(x)=1+2x4,

f(0)=1+2(0)4=14=1

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

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

f(x)≈f(0)+f′(0)(x−0)≈f(0)+f′(0)(x)

Substitute
f′(0)=12 and
f(0)=1,

f(x)≈1+12(x−0)=1+x2

Here, the function
f(x)=1+2x4,

1+2x4≈1+12x

Hence the required result is verified.

Now to determine the value of *x* for which
|f(x)−L(x)|<0.1.

Since the linear approximation at
a=0 is
1+2x4≈1+12x, the function
f(x)=1+2x4 and the linearization
L(x)=1+12x.

|1+2x4−(1+12x)|<0.1−0.1<1+2x4−1−12x<0.11−0.1<1+2x4−12x<1+0.10.9<1+2x4−12x<1.1

Here,
y=1+2x4−1+12x.

Use the online graphing calculator to draw
y=1+2x4−1+12x zoom towards the value
y=0.9 and y=1.1 as shown below in Figure 1.

From Figure 1, it is observed that the graph of the function
y=1+2x4−12x lies between 0.9 and 1.1 whenever
−0.3689<x<0.6777.

That is, the value of
−0.3689<x<0.6777 for which the linear approximation of function
1+2x4 is accurate to within 0.1.