#### To determine

**(a)**

**To Find:**

Linearization of the function near 3

#### Answer

Lx=6.25-3x4

#### Explanation

**Concept:**

Use of Linear Approximations and Differentials, Derivative of function

**Formula:**

(i) Linearization Formula:

Lx=fa+f'a(x-a)

(ii) Power rule:

ddxxn=n xn-1

(iii) Chain Rule:

dydx= dydududx

(iv) Use distributive property:

a (b+ c) = a*b + a*c

(v) Quadratic Formula:

x=-b±b2-4ac2a

**Given:**

fx= 25-x2, a=3

**Calculation:**

Consider,

fx= 25-x2

fx=(25-x2)12

Let, u= 25-x2

Differentiating u with respect to x

dudx=-2x

So, function becomes,

fx=u1/2

Now, differentiating with respect to x.

ddxfx=ddxu1/2

By using power rule and Chain Rule

f'(x) =12u12-1dudx

f'(x) =12u-12dudx

Simplifying and substituting the value of u and

dudx

f'(x)=1225-x2-12(-2x)

f'x=-x 25-x2-12

So, value of fx and f'x at a=3 is f3 and f'(3)

f3= 25-32

f3= 25-9

f3=16

f3=4

f'(3)=-(3) 25-32-12

f'(3)=-(3) 25-9-12

f'(3)=-(3) 16-12

f'(3)=-34

Now, linearization at a is

By Linearization Formula

Lx=fa+f'a(x-a)

∴ linearization at a=3 is

Lx=f3+f'3(x-3)

Lx=4+-34(x-3)

By using distributive law

Lx=4-34x+ 94

Lx=4-34x+ 2.25

Lx=6.25-34x

**Conclusion:**

Linearization Function is Lx=6.25-34x

#### To determine

**(b)**

**To illustrate:**

Linearization of function f=25-x2 near 3 is Lx=6.25-34x

by graphing function and linear approximation

#### Answer

Linearization of function f=25-x2 near 3 is Lx=6.25-34x

#### Explanation

**Concept:**

By using linear approximation

**Given:**

i. f=25-x2

ii. Lx=6.25-34x

**Graph:**

Graph of f=25-x2 and linearization function at x = 3

Function and line touches at point (3, 4)

**Conclusion:**

Linearization of function f=25-x2 near 3 is Lx=6.25-34x

#### To determine

**(c)**

**To find:**

The value of x for which the approximation is good

#### Answer

Value of x lies between 2.24 to 3.36

#### Explanation

**Concept:**

Use of Linear Approximations and Differentials, Derivative of function

**Formula:**

(i) Quadratic Formula:

x=-b±b2-4ac2a

**Given:**

gx=6.25-34x

f(x)=25-x2

**Calculation:**

gx=6.25-34x

and f(x)=25-x2

Now, gx≥fx(refer graph)

Suppose gx-fx=0.1 then

∴ 6.25-34x- 25-x2=0.1

∴ 6.25-34x- 0.1=25-x2

∴6.15-34x=25-x2

Squaring on both sides

∴6.15-34x2=25-x22

∴37.8225+916x2-36.94x=25-x2

∴916x2+x2-36.94x+37.8225-25=0

∴2516x2-36.94x+12.8225=0

Comparing the given equation with ax2+bx+c=0 to get values of a, b and c

a=2516, b=-36.94, c=12.8225

By using quadratic formula,

x=-b±b2-4ac2a

x=--36.94± -36.942-4 251612.82252 2516

x=9.2± 85.1-80.13.13

x=9.2± 53.13

x=3.65 and 2.22

So it follows that between these values the difference between two functions is less than 0.1. So when value of x lies between 2.24 to 3.66 the linear approximation is within 0.1 of original function.

**Conclusion:**

The value of x lies between 2.24 to 3.66