Explanation Given Information: The provided function is, f ( x ) = x 4 − 2 x 3 + x 2 − 2 x + 1 Graph: Consider the provided function, f ( x ) = x 4 − 2 x 3 + x 2 − 2 x + 1 Now, differentiate the provided function with respect to x as, f ′ ( x ) = d d x ( x 4 − 2 x 3 + x 2 − 2 x + 1 ) = d d x ( x 4 ) − d d x ( 2 x 3 ) + d d x ( x 2 ) − d d x ( 2 x ) + d d x ( 1 ) = 4 x 3 − 6 x 2 + 2 x − 2 Now, plot this graph using TI-83 as, Step1: Press ON key. Step2: Press “Y=” key. Step3: Enter the equation as below, Y 1 = 4 × X ^ 3 − 6 × X ^ 2 + 2 × X − 2 Step4: Press GRAPH key. The graph so obtained is depicted below, Now, as it can be observed from the graph that y = 0 or f ' ( x ) = 0 for x ≈ 1.40 . Now, to find the minima, substitute x = 1.40 in the equation, f ( x ) = x 4 − 2 x 3 + x 2 − 2 x + 1 . f ( 1.40 ) = ( 1.40 ) 4 − 2 ( 1.40 ) 3 + ( 1.40 ) 2 − 2 ( 1.40 ) + 1 = 3.8416 − 5.488 + 1.96 − 2.80 + 1 = − 1.49 Therefore, f ( x ) has absolute minimum at ( 1.40 , − 1.49 ) . Now, obtain f ″ ( x ) as, f ″ ( x ) = d d x ( 4 x 3 − 6 x 2 + 2 x − 2 ) = d d x ( 4 x 3 ) − d d x ( 6 x 2 ) + d d x ( 2 x ) − d d x ( 2 ) = 12 x 2 − 12 x + 2 Now, plot this graph using TI-83 as, Step1: Press ON key

7th Edition

ISBN: 9781337291248

Author: Waner, Stefan.

Publisher: Cengage Learning,

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Chapter 5.4, Problem 27E

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To graph: The function f(x)=x4−2x3+x2−2x+1 and label point of inflection and all the relative and absolute extrema and vertical and horizontal asymptotes.

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Chapter 5.4, Problem 26E

Chapter 5.4, Problem 28E

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Chapter 5 Solutions

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To graph: The function f ( x ) = x 4 − 2 x 3 + x 2 − 2 x + 1 and label point of inflection and all the relative and absolute extrema and vertical and horizontal asymptotes.

Solution Summary: The author explains how to plot a graph using TI-83.

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To graph: The function f(x)=x4−2x3+x2−2x+1 and label point of inflection and all the relative and absolute extrema and vertical and horizontal asymptotes.

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