Explanation Given: ( 1.02 ) 0.7 and ( 1.02 ) − 0.3 Formula used: Linear approximation of Δ f : If f is differentiable at x = a and Δ x is small then Δ f ≈ f ′ ( a ) Δ x Where Δ f = f ( a + Δ x ) − f ( a ) . Calculation: Let the function f ( x ) = ( 1 + x ) k and a = 0 . The linear approximation is f ( a + Δ x ) − f ( a ) = f ′ ( a ) Δ x Differentiate with respect to x . f ′ ( x ) = k ( 1 + x ) k − 1 Substitute 0 for x in the above derivative. f ′ ( 0 ) = k ( 1 + 0 ) k − 1 = k By the linear approximation, f ( 0 + Δ x ) − f ( 0 ) ≈ f ′ ( 0 ) Δ x ( 1 + Δ x ) k − 1 ≈ k Δ x ( 1 + Δ x ) k ≈ 1 + k Δ x Now, find the value of ( 1

ACHIEVE FOR CALCULUS 1 TERM >CSI CUSTOM<

4th Edition

ISBN: 9781319411657

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.1, Problem 72E

To determine

Show that for any real number k,(1+Δx)k≈1+kΔx for small Δx. Also, estimate (1.02)0.7 and (1.02)−0.3

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Chapter 4.1, Problem 71E

Chapter 4.1, Problem 73E

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ACHIEVE FOR CALCULUS 1 TERM >CSI CUSTOM<

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Show that for any real number k , ( 1 + Δ x ) k ≈ 1 + k Δ x for small Δ x . Also, estimate ( 1.02 ) 0.7 and ( 1.02 ) − 0.3

Solution Summary: The author explains the linear approximation of Delta f.

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Show that for any real number k,(1+Δx)k≈1+kΔx for small Δx. Also, estimate (1.02)0.7 and (1.02)−0.3

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