Explanation Given information: A function is given as, y = sec 2 x − 2 tan x on interval [ − π 6 , π 3 ] . Calculation: A function is given as, y = sec 2 x − 2 tan x on interval [ − π 6 , π 3 ] . Now let us first find the critical points. y = sec 2 x − 2 tan x ⇒ y ' = 2 sec 2 x tan x − 2 sec 2 x ⇒ y ' = 2 sec 2 x ( tan x − 1 ) At critical points, y ' = 0 ⇒ 2 sec 2 x ( tan x − 1 ) = 0 ⇒ sec x = 0 ⇒ This is not possible . and tan x − 1 = 0 ⇒ tan x = 1 ⇒ x = π 4 Now calculating function values at the critical points and extreme points of interval [ − π 6 , π 3 ]

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.2, Problem 52E

To determine

The maximum and minimum values of the function for the given interval.

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Chapter 4.2, Problem 51E

Chapter 4.2, Problem 53E

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Taylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52

Could you explain this using the formula I attached and polar coorindates

Could you explain this using the formula I attached and polar coordinates

Chapter 4 Solutions

CALCULUS (CLOTH)

Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E

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Chapter 4 Solutions