Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises: 55

Answer

At 2 points $A(1,-6)$ and $B(-2,21)$, the tangent line to the curve is horizontal.

Work Step by Step

$$y=f(x)=2x^3+3x^2-12x+1$$ 1) First, find the derivative of function $f(x)$ $$y'=\frac{df(x)}{dx}=2\frac{d}{dx}(x^3)+3\frac{d}{dx}(x^2)-12\frac{d}{dx}(x)+\frac{d}{dx}(1)$$ $$y'=6x^2+6x-12$$ $$y'=6(x^2+x-2)$$ $$y'=6(x-1)(x+2)$$ 2) The slope of the tangent line of the curve $y$ at a point $A (a,b)$ would be the value of $y'(a)$. A tangent line that is horizontal at point $A(a,b)$ is one that has a slope value of $0$, or, in other words, $y'(a)=0$ $$y'(a)=0$$ $$6(a-1)(a+2)=0$$ $$a=1$$ or $$a=-2$$ That means at 2 points $A(1,b)$ and $B(-2,d)$, the tangent line to the curve is horizontal. Now, these 2 points $A$ and $B$ must lie in the curve $y=f(x)$. Therefore, to find $b$ and $d$, we would apply the function equation of the curve: $$b=2\times1^3+3\times1^2-12\times1+1=-6$$ $$d=2\times(-2)^3+3\times(-2)^2-12\times(-2)+1=21$$ Therefore, at 2 points $A(1,-6)$ and $B(-2,21)$, the tangent line to the curve is horizontal.
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