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: 19


The slope of the curve at $x=1$ is $-1$.

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 $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$f(x)=5x-3x^2\hspace{1cm}x=1$$ To calculate the slope of the curve at $x=1$, we apply the above formula: $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ Here $a=x=1$ $$m=\lim_{h\to0}\frac{f(h+1)-f(1)}{h}=\lim_{h\to0}\frac{\Big(5(h+1)-3(h+1)^2\Big)-(5\times1-3\times1^2)}{h}$$ $$m=\lim_{h\to0}\frac{\Big(5h+5-3(h^2+2h+1)\Big)-2}{h}$$ $$m=\lim_{h\to0}\frac{5h+5-3h^2-6h-3-2}{h}=\lim_{h\to0}\frac{-3h^2-h}{h}$$ $$m=\lim_{h\to0}(-3h-1)$$ $$m=-3\times0-1=-1$$
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