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: 69

Answer

The function in need to find is $y=\frac{3}{16}x^3-\frac{9}{4}x+3$

Work Step by Step

$$y=ax^3+bx^2+cx+d$$ 1) Find $y'$ $$y'=3ax^2+2bx+c$$ 2) The slope of the tangent line $(l)$ to the curve $f$ at point $A(a,b)$ equals $y'(a)$. Horizontal tangents have slope 0. That means $y'(a)=0$. To have horizontal tangent at $(-2,6)$ means $$y'(-2)=0$$ $$3a\times(-2)^2+2b\times(-2)+c=0$$ $$12a-4b+c=0\hspace{0.5cm}(1)$$ Similarly, to have horizontal tangent at $(2,0)$ means $$y'(2)=0$$ $$3a\times2^2+2b\times2+c=0$$ $$12a+4b+c=0\hspace{0.5cm}(2)$$ We subtract (2) from (1) and have $$-8b=0$$ $$b=0$$ So, function $y$ now becomes $$y=ax^3+cx+d$$ 3) Since there are tangent lines at $(-2, 6)$ and $(2,0)$, these 2 points also lie in the graph of function $y$. That means $(-2)^3a+(-2)c+d=6$. Or $-8a-2c+d=6\hspace{0.5cm}(3)$ And also $2^3a+2c+d=0$. Or $8a+2c+d=0\hspace{0.5cm}(4)$. Combing $(1), (3), (4)$, we have 3 equations to find 3 constant $a, c, d$. Solving them, we find that $a=\frac{3}{16}$, $c=-\frac{9}{4}$, $d=3$. Therefore, the function in need to find is $y=\frac{3}{16}x^3-\frac{9}{4}x+3$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.