Explanation Let f ( x ) = | x − 1 | + | 2 x − 6 | = { 7 - 3 x , x < 1 5 − x , 1 ≤ x < 3. 3 x − 7 , x ≥ 3 The derivation of f ( x ) is never zero but does not exist at the transition points x = 1 and x = 3 .the critical points of f are x = 1 and x = 3

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4, Problem 37CRE

To determine

To find out the critical points and extreme values of f(x)=|x−1|+|2x−6| in [0,8].

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Chapter 4, Problem 36CRE

Chapter 4, Problem 38CRE

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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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To find out the critical points and extreme values of f ( x ) = | x − 1 | + | 2 x − 6 | in [ 0 , 8 ] .

Solution Summary: The author explains that the derivation of f(x) is never zero but does not exist at the transition points.

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To find out the critical points and extreme values of f(x)=|x−1|+|2x−6| in [0,8].

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Chapter 4 Solutions