University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 118: 15


The equation of the tangent line is $y=12x-16$.

Work Step by Step

The slope $m$ of the tangent line of the curve $f(x)$ at $A(a,b)$ is also the slope of the curve at that point, which is calculated by $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$f(t)=t^3\hspace{1cm}A(2,8)$$ 1) Find the slope $m$ of the tangent: $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ Here $a=2$ and $g(a)=b=8$ $$m=\lim_{h\to0}\frac{f(h+2)-8}{h}=\lim_{h\to0}\frac{(h+2)^3-8}{h}=\lim_{h\to0}\frac{h^3+6h^2+12h+8-8}{h}$$ $$m=\lim_{h\to0}\frac{h^3+6h^2+12h}{h}=\lim_{h\to0}(h^2+6h+12)$$ $$m=0^2+6\times0+12=12$$ 2) Find the equation of the tangent line at $A(2,8)$: The tangent line would have this form: $$y=12x+m$$ Substitute $A(2,8)$ here to find $m$: $$12\times2+m=8$$ $$24+m=8$$ $$m=-16$$ So the equation of the tangent line is $y=12x-16$.
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