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: 11


The equation of the tangent line is $y=4x-3$.

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 $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$y=f(x)=x^2+1\hspace{1cm}A(2,5)$$ 1) Find the slope $m$ of the tangent: $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ Here $a=2$ and $f(a)=b=5$ $$m=\lim_{h\to0}\frac{f(2+h)-5}{h}=\lim_{h\to0}\frac{(2+h)^2+1-5}{h}$$ $$m=\lim_{h\to0}\frac{4+4h+h^2+1-5}{h}=\lim_{h\to0}\frac{4h+h^2}{h}$$ $$m=\lim_{h\to0}(4+h)$$ $$m=4+0=4$$ 2) Find the equation of the tangent line at $A(2,5)$: The tangent line would have this form: $$y=4x+m$$ Substitute $A(2,5)$ here to find $m$: $$4\times2+m=5$$ $$8+m=5$$ $$m=-3$$ So the equation of the tangent line is $y=4x-3$.
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