Linearization

5.5 Linearization and Differentials

Linearization

Multiple Choice

The linearization of function y=f(x) near the value x=a is given by:

• L(x) = f(a) + f'(a) (x - a)

• L(a) = f'(x) + f(x) (x - a)

Multiple Choice

f(x)= cos x , $$a=\frac{\pi}{2}$$ , approximate cos 1.75 without calculator

• 0.345

• 0.9998766767

• -0.1782460556

• -0.8933566678

Multiple Choice

approximate $$\sqrt[]{1.02}$$ without calculator

• 1.0004

• 0.98

• 1

• 1.01

Multiple Choice

Use linearization to approximate $$\sqrt[]{123}$$

• 11.234

• 11.090909

• 11

• 31

Multiple Choice

Use Linearization to approximate $$\sqrt[3]{26}$$

• 4.973189

• 7

• 4

• 2.962

Multiple Choice

Let L(x) = 3(x - 4) - 2.  Find L(3.9).

• -1.7

• -2.3

​Differentials

Multiple Choice

y=x⁵ +37x, x=1 , dx=0.001

• 0.42

• -0.02

• -0.05

Multiple Choice

y=sin 3x , x=Pi , dx=-0.02

• 0.42

• 0.06

• -0.05

Multiple Choice

Let y = x³

Find the differential when x =2 and dx = 0.1.

• 1.2

• -9.02

Multiple Choice

Find dy if dy/dx = 7 and dx = 1/2. 

• 7/2

• 14

​Newton's Method

Multiple Choice

Explain the concept of Newton's method to approximate zeros of a function.

• Newton's method is an iterative method for finding the approximate zeros of a real-valued function.

• Newton's method is used to calculate the area under a curve

Multiple Choice

What is the formula to update the guess in Newton's method?

• x₁= x₀ * f(x₀) / f'(x₀)

• x₁= x₀ - f(x₀) / f'(x₀)

Multiple Select

Use Newton's Method to solve x³ +3x+1=0 , x₀ = -0.3

• x₁ =-0.4

• x₁ =-0.322324159

• x₂ =-0.3221853603

• x₂ = -0.4562212

Multiple Select

Use Newton's method to solve x³ +4x² -x-1=0 , x₀=-1

• x₁=-0.5

• x₁ =0.75

• x₂ =-0.411767

• x₂ =-9.876

Multiple Choice

When does Newton's method fail to find an approximate zero of a function?

• Vertical tangent or distant initial guess

• Linear function or close initial guess