Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.1 Introducing the Derivative - 3.1 Execises: 60

Answer

This expression represents the slope of the tangent of the function $f(x)=3x^2+4x$ at $a=1$ and its value is $m_{tan}=10$.

Work Step by Step

The value of the slope of the tangent of the function $f(x)$ at $a$ is given by $$m_{tan}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$$ We see that if we take $a=1$ and $f(x)=3x^2+4x$ (we also calculate $f(1)=3\cdot1^2+4\cdot1=7$) we see that the given expression gives the slope of the tangent of $f(x)=3x^2+4x$ at $a=1$. Let us calculate this slope $$m_{tan}=\lim_{x\to1}\frac{3x^2+4x-7}{x-1}.$$ Since we have the expression of the form $0/0$ let us factor the expression in the numerator. First find its zeros: $$x_{1,2}=\frac{-4\pm\sqrt{4^2+84}}{2\cdot3}=\frac{-4\pm 10}{6}=\left\{^1_{-7/3} \right..$$ We know that if we have the function $\alpha x^2+\beta x+\gamma$ with zeroes $x_1$ and $x_2$ it can be factored as $\alpha x^2+\beta x+\gamma=\alpha(x-x_1)(x-x_2)$. Applying this to the function in the numerator of the limit we get $3x^2+4x-7=3(x-1)(x+7/3)=(x-1)(3x+7)$ so putting this into the numerator we have $$m_{tan}=\lim_{x\to1}\frac{(x-1)(3x+7)}{x-1}=\lim_{x\to1}(3x+7)=3\cdot1+7=10.$$
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