Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Problems Plus - Problems - Page 271: 2

Answer

The curves are tangent to each other at the point $(2,6)$ We can see a sketch of the graphs below.

Work Step by Step

$y = x^3-3x+4$ $\frac{dy}{dx} = 3x^2-3$ $y = 3(x^2-x) = 3x^2-3x$ $\frac{dy}{dx} = 6x-3$ If the two curves are tangent to each other at a point, then the slopes of the graphs must be equal at that point. We can find the points where the slopes are equal. $3x^2-3 = 6x-3$ $3x^2-6x = 0$ $3x(x-2) = 0$ $x=0$ or $x=2$ When $x=0$: $x^3-3x+4 = (0)^3-3(0)+4 = 4$ $3(x^2-x)= 3(0^2-0) = 0$ Since the values are not equal at $x=0$, the curves are not tangent at $x=0$ When $x=2$: $x^3-3x+4 = (2)^3-3(2)+4 = 6$ $3(x^2-x)= 3(2^2-2) = 6$ Since the values are equal at $x=2$, and the slopes are equal at $x=2$, the curves are tangent at $x=2$. Therefore, the curves are tangent to each other at the point $(2,6)$. We can see a sketch of the graphs below.
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