Explanation Given data: y The graph of f ′ is shown in figure here, and given value of function f is f ( − 1 ) = 2 . Method/Formula used: If f is antiderivative of the function f ′ then by Fundamental Theorem f ( x ) = ∫ f ′ ( x ) d x + C Calculation: The graph (parabola) of the function shows that f ′ is quadratic function whose roots are 2.5 and − 0.5 . Therefore f ′ ( x ) = ( x − 2.5 ) ( x − 0.5 ) = x 2 − 3 x + 1.25 . Now, f ( x ) = ∫ f ′ ( x ) d x + C , where C is a constant. ⇒ f ( x ) = ∫ ( x 2 − 3 x + 1.25 ) d x + C ⇒ f ( x ) = x 3 3 − 3 x 2 2 + 1.25 x + C ..... (1) We have f ( − 1 ) = 2 ..... (2) From results (1) and (2) f ( − 1 ) = ( − 1 ) 3 3 − 3 ( − 1 ) 2 2 + 1

14th Edition

ISBN: 9781464151088

Author: Associate Professor Laura Taalman, Peter Kohn

Publisher: W.H. Freeman & Company

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Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

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Chapter 4.5, Problem 12C

To determine

To approximate:The valuefunction f(1) using graph of f′ and given value of f is given, an area approximation and Fundamental Theorem.

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Chapter 4.5, Problem 11C

Chapter 4.5, Problem 13C

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To approximate: The valuefunction f ( 1 ) using graph of f ′ and given value of f is given, an area approximation and Fundamental Theorem.

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To approximate:The valuefunction f(1) using graph of f′ and given value of f is given, an area approximation and Fundamental Theorem.

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