Explanation Given: f ( x ) = ( x 2 + 1 ) ( 1 − x 3 ) , where x = 1 Concept used: d ( x n ) d x = n ⋅ x n − 1 & d ( c ) d x = 0 , c ∈ R Equation of the tangent line passing through the point (a, b) with slope m is y − b = m ( x − a ) Calculation: Slope of the tangent is the derivative of the given function at the given point. So differentiating f(x) with respect to x, we get f ′ ( x ) = d ( x 2 + 1 ) d x ⋅ ( 1 − x 3 ) + d ( 1 − x 3 ) d x ⋅ ( x 2 + 1 ) = 2 x ( 1 − x 3 ) − 3 x 2 ( x 2 + 1 )

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 3.2, Problem 29PS

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To find the equation of the tangent line of the given function at the given point.

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Chapter 3.2, Problem 28PS

Chapter 3.2, Problem 30PS

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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 3 Solutions

Calculus

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To find the equation of the tangent line of the given function at the given point.

Solution Summary: The author explains the equation of the tangent line passing through the point (a, b) with slope m.

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Chapter 3 Solutions