Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises - Page 142: 79

Answer

Given, $y=6x^3+5x-3$. A tangent line with slope $4$ must be drawn at a point on the curve, which has the same slope, that is, $4$. For finding the slope of $y$, let's differentiate it with respect to $x$, $$\frac{dy}{dx} = 18x^2+5$$This slope needs to be equal to $4$. Thus, $$18x^2+5=4\implies18x^2+1=0$$$$x^2=-\frac{1}{18}$$No real value satisfies the above equation, hence the given curve has no tangent line with slope $=4$.

Work Step by Step

Given, $y=6x^3+5x-3$. A tangent line with slope $4$ must be drawn at a point on the curve, which has the same slope, that is, $4$. For finding the slope of $y$, let's differentiate it with respect to $x$, $$\frac{dy}{dx} = 18x^2+5$$This slope needs to be equal to $4$. Thus, $$18x^2+5=4\implies18x^2+1=0$$$$x^2=-\frac{1}{18}$$No real value satisfies the above equation, hence the given curve has no tangent line with slope $=4$.
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