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 - Page 182: 79


$c=-\frac{1}{3}$ is the solution to the exercise.

Work Step by Step

$$(l): y=2x+3$$ The slope of line $(l)$ is $2$. $$(S): y=f(x)=cx^2$$ - Derivative of $f(x)$: $$f'(x)=2cx$$ Line $(l)$ can only be tangent with curve $(S)$ if both of the followings happen: 1) Line $(l)$ intersects curve $(S)$ at at least one point $A(a, b)$. 2) At point A, $f'(a)$ equals the slope of line $(l)$. *Consider 1): If $A(a,b)$ is where line $(l)$ and curve $(S)$ intersects, then $A$ must lie in both the line and the curve. Therefore, $$b=ca^2\hspace{1cm}(1)$$ and $$b=2a+3\hspace{1cm}(2)$$ Consider 2): At point $A(a,b)$:$$f'(a)=2$$ $$2ca=2$$ $$ca=1$$ $$c=\frac{1}{a}\hspace{1cm}(3)$$ Substitute (3) into (1), we have $$b=\frac{1}{a}a^2=a\hspace{1cm}(4)$$ Substitue (4) into (2), we have $$a=2a+3$$ $$a=-3$$ Substitute $a=-3$ to (3), we have $$c=-\frac{1}{3}$$ $c=-\frac{1}{3}$ is the result.
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