Explanation Given information: The function, f ( x ) = x x 2 + 1 . Formula used: The quotient rule of differentiation d d x [ f ( x ) g ( x ) ] = g ( x ) f ' ( x ) − f ( x ) g ' ( x ) [ g ( x ) ] 2 Let a function g be continuous on closed interval [ c , d ] and differentiable on open interval ( c , d ) , If first derivative of the function is greater than zero that is g ' ( x ) > 0 for every xin ( c , d ) , then the function g is increasing on interval [ c , d ] . If first derivative of the function is less than zero that is g ' ( x ) < 0 for every xin ( c , d ) , then the function g is decreasing on interval [ c , d ] . Calculation: Consider the function, f ( x ) = x x 2 + 1 . Differentiate the function with respect to x , recall the quotient rule of differentiation d d x [ f ( x ) g ( x ) ] = g ( x ) f ' ( x ) − f ( x ) g ' ( x ) [ g ( x ) ] 2 . Apply it, d d x [ x x 2 + 1 ] = ( x 2 + 1 ) x ' − x ( x 2 + 1 ) ' [ x 2 + 1 ] 2 = x 2 + 1 − x ( 2 x ) [ x 2 + 1 ] 2 = − x 2 + 1 [ x 2 + 1 ] 2 Equate the derivative obtained above to 0 to get the critical number for the function, − x 2 + 1 [ x 2 + 1 ] 2 = 0 − x 2 + 1 = 0 x 2 = 1 x = ± 1 That is, either x = 1 or x = − 1 . Therefore, critical point are x = 1 and x = − 1 . Since the function is polynomial, construct the open intervals to test the increasing and decreasing nature of the function, ( − ∞ , − 1 ) , ( − 1 , 1 ) and ( 1 , ∞ ) . Observe that on the above three intervals, the derivative function has no zeros, and is continuous. Recall if a function g is continuous on closed interval [ c , d ] and differentiable on open interval ( c , d ) , If first derivative of the function is greater than zero that is g ' ( x ) > 0 for every xin ( c , d ) , then the function g is increasing on interval [ c , d ] . If first derivative of the function is less than zero that is g ' ( x ) < 0 for every xin ( c , d ) , then the function g is decreasing on interval [ c , d ] . Consider the interval, ( − ∞ , − 1 ) . Choose any number that lies in between the interval, say k = − 3 . Apply it, substitute k = − 3 , in the f ' ( x ) and check the value, f ' ( − 3 ) = − ( − 3 ) 2 + 1 [ ( − 3 ) 2 + 1 ] 2 = − 9 + 1 ( 10 ) 2 = − 0

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ISBN: 9781429241861

Author: Laura Taalman, Peter Kohn

Publisher: W. H. Freeman

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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 3.1, Problem 1TF

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To calculate: For the function, f(x)=xx2+1 , the intervals in which the first derivative of the function is positive and negative also construct a sign chart for the same and mark the x values where derivative is zero or undefined.

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Chapter 3.1, Problem 69P

Chapter 3.1, Problem 2TF

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Chapter 3 Solutions

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For the function, f ( x ) = x x 2 + 1 , the intervals in which the first derivative of the function is positive and negative also construct a sign chart for the same and mark the x values where derivative is zero or undefined.

Solution Summary: The author calculates the function f(x)=x 2+1 based on the intervals in which the first derivative is positive and negative.

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To calculate: For the function, f(x)=xx2+1 , the intervals in which the first derivative of the function is positive and negative also construct a sign chart for the same and mark the x values where derivative is zero or undefined.

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Chapter 3 Solutions