Explanation Given: Function f ( x ) = 2 x − 5 x + 1 . Formula Used: Equation of line using slope-point form is given as ( y − y 1 ) = m ( x − x 1 ) where ( x 1 , y 1 ) is point of line and m is the slope of the line Calculation: Function is given as f ( x ) = 2 x − 5 x + 1 . Let us find the point on graph where tangent is drawn Put x = 0 in f ( x ) , we get f ( x ) = 2 x − 5 x + 1 f ( 0 ) = − 5 1 f ( 0 ) = − 5 Also, slope of the line is calculated by taking the derivative of the function f ′ ( x ) = d d x ( 2 x − 5 x + 1 ) Using quotient rule of derivative, we have f ′ ( x ) = d d x ( 2 x − 5 ) × ( x + 1 ) − ( 2 x − 5 ) × d d x ( x + 1 ) ( x + 1 ) 2 f ′ ( x ) = (

6th Edition

ISBN: 9781119399322

Author: Hughes-Hallett

Publisher: WILEY

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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 3.4, Problem 34P

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To find the equation of tangent line to the graph of f(x)=2x−5x+1 at the point at which x=0.

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Chapter 3.4, Problem 33P

Chapter 3.4, Problem 35P

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To find the equation of tangent line to the graph of f ( x ) = 2 x − 5 x + 1 at the point at which x = 0 .

Solution Summary: The author explains how to find the equation of tangent line to the graph of f(x)=2x-5x+1.

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To find the equation of tangent line to the graph of f(x)=2x−5x+1 at the point at which x=0.

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Chapter 3 Solutions