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: 16


The equation of the tangent line is $y=6x-2$.

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 $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$f(t)=t^3+3t\hspace{1cm}A(1,4)$$ 1) Find the slope $m$ of the tangent: $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ Here $a=1$ and $f(a)=b=4$ $$m=\lim_{h\to0}\frac{f(h+1)-4}{h}=\lim_{h\to0}\frac{(h+1)^3+3(h+1)-4}{h}$$ $$m=\lim_{h\to0}\frac{h^3+3h^2+3h+1+3h+3-4}{h}$$ $$m=\lim_{h\to0}\frac{h^3+3h^2+6h}{h}=h^2+3h+6$$ $$m=0^2+3\times0+6=6$$ 2) Find the equation of the tangent line at $A(1,4)$: The tangent line would have this form: $$y=6x+m$$ Substitute $A(1,4)$ here to find $m$: $$6\times1+m=4$$ $$6+m=4$$ $$m=-2$$ So the equation of the tangent line is $y=6x-2$.
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