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: 77

Answer

When $a=-\frac{1}{2}$ and $b=2$, the requirement of the exercise is satisfied.

Work Step by Step

$$(l): 2x+y=b$$ $$(l): y=-2x+b$$ The slope of line $(l)$ is $-2$. $$(S): y=f(x)=ax^2$$ - Derivative of $f(x)$: $$f'(x)=2ax$$ Line $(l)$ can only be tangent with parabola $(S)$ when $x=2$ in the case that $f'(2)$ equals the slope of line $(l)$. In other words, $$f'(2)=-2$$ $$2a\times2=-2$$ $$4a=-2$$ $$a=-\frac{1}{2}$$ Therefore, the equation of parabola $(S)$ is $$(S):y=f(x)=-\frac{1}{2}x^2$$ - According to the equation of parabola $(S)$, when $x=2$, $y=-\frac{1}{2}\times2^2=-2$ So, point $A(2,-2)$ lies in parabola $(S)$. However, since point $A$ is where tangent line $(l)$ touches parabola $(S)$, $A$ also lies in line $(l)$, which means $$2\times2-2=b$$ $$b=2$$ In conclusion, $a=-\frac{1}{2}$ and $b=2$ is the solution to this exercise.
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