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: 20


The slope of the curve at $x=-2$ is $10$.

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 $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$f(x)=x^3-2x+7\hspace{1cm}x=-2$$ To calculate the slope of the curve at $x=-2$, we apply the above formula: $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ Here $a=x=-2$ $$m=\lim_{h\to0}\frac{f(h-2)-f(-2)}{h}$$ $$m=\lim_{h\to0}\frac{\Big((h-2)^3-2(h-2)+7\Big)-\Big((-2)^3-2(-2)+7\Big)}{h}$$ $$m=\lim_{h\to0}\frac{\Big(h^3-6h^2+12h-8-2h+4+7\Big)-\Big(-8+4+7\Big)}{h}$$ $$m=\lim_{h\to0}\frac{h^3-6h^2+10h}{h}=\lim_{h\to0}(h^2-6h+10)$$ $$m=0^2-6\times0+10=10$$
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