Explanation Given Information: The provided function is f ( x ) = ( x 2 + 1 ) ( x 3 + x ) and x = 1 . Formula used: Slope of tangent to the graph of the function f ( x ) at point ( a , b ) is given by f ′ ( a ) . Power rule of a function f ( x ) = x n is f ′ ( x ) = n x n − 1 , where n is a constant. The derivative of constant is 0 . Sum rule of derivative is [ f + g ] ′ ( x ) = f ′ ( x ) + g ′ ( x ) , where, f ( x ) and g ( x ) are any two differentiable functions. Equation of the line is y = m x + b where, m is the slope and b = y 1 − m x 1 when line passes through ( x 1 , y 1 ) . Calculation: Consider the provided function f ( x ) = ( x 2 + 1 ) ( x 3 + x ) , To find the slope of the tangent to the graph f ( x ) = ( x 2 + 1 ) ( x 3 + x ) by determining derivative of the function f ( x ) . Apply product rule to the function, f ′ ( x ) = [ d d x ( x 2 + 1 ) ⋅ ( x 3 + x ) + ( x 2 + 1 ) ⋅ d d x ( x 3 + x ) ] To calculate the derivative, apply sum rule of derivative, f ′ ( x ) = ( x 3 + x ) [ d d x ( x 2 ) + d d x ( 1 ) ] + ( x 2 + 1 ) [ d d x ( x 3 ) + d d x ( x ) ] Now, apply product rule and derivative of constant rule, f ′ ( x ) = [ ( x 3 + x ) ⋅ { 2 x 2 − 1 + 0 } + ( x 2 + 1 ) ⋅ { 3 x 3 − 1 + 1 x 1 − 1 } ] = [ ( x 3 + x ) ( 2 x ) + ( x 2 + 1 ) ( 3 x 2 + 1 ) ] = [ 2

Finite Mathematics and Applied Calculus (MindTap Course List)

7th Edition

ISBN: 9781337274203

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 11.3, Problem 71E

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To calculate: The equation of the tangent to the graph of the function f(x)=(x2+1)(x3+x) at x=1.

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Chapter 11.3, Problem 70E

Chapter 11.3, Problem 72E

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The equation of the tangent to the graph of the function f ( x ) = ( x 2 + 1 ) ( x 3 + x ) at x = 1 .

Solution Summary: The author calculates the equation of tangent to the graph of the function f(x) at x=1.

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To calculate: The equation of the tangent to the graph of the function f(x)=(x2+1)(x3+x) at x=1.

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