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 - Page 133: 10


$a.$ the slope is $m_{tan} = -11.$ $b.$ the equation is $y=-11x+4.$ $c.$ the graph is on the figure below.

Work Step by Step

$a.$ Using the expression for the slope of the tangent line given in definition (1) with $a=1$ (the first coordinate of the point $P(1,-7)$) we have $$m_{tan} = \lim_{x\to 1}\frac{f(x)-f(1)}{x-1} = \lim_{x\to 1}\frac{-3x^2-5x+1 - (-3\cdot1^2-5\cdot1+1)}{x-1} =\lim_{x\to 1}\frac{-3x^2-5x+1 +7}{x-1} = \lim_{x\to 1}\frac{-3x^2-5x+8}{x-1}.$$ To find this limit we will factorize the quadratic function in the numerator. For its zeros by the formula for zeros of the quadratic function we have $$x_{1,2}=\frac{-(-5)\pm\sqrt{(-5)^2-4\cdot(-3)\cdot8}}{2\times(-3)} = \frac{5\pm 11}{-6} = \left\{^{-8/3}_1\right.$$ Now we know that every quadratic function of the form of $a_1x^2+a_2x+a_3$ with zeros $x_1$ and $x_2$ can be factored as $$a_1x^2+a_2x+a_3=a_1(x-x_1)(x-x_2)$$ which in our case gives $$-3x^2-5x+8 = -3(x+8/3)(x-1) = -(3x+8)(x-1).$$ Now returning to solving the limit we have $$m_{tan} = \lim_{x\to 1}\frac{-3x^2-5x+8}{x-1} =\lim_{x\to 1} \frac{-(3x+8)(x-1)}{x-1} = \lim_{x\to 1}(-(3x+8))=-11.$$ $b.$ Using the formula fro the equation of the tangent $y-f(a)=m_{tan}(x-a)$ for $a=1$ and $f(a)=-7$ (the second coordinate of point $P(1,-7)$) we have $$y-(-7) = -11(x-1)\Rightarrow y+7=-11x+11$$ which gives $$y=-11x+4.$$ $c.$ The graph is on the figure below. The function is solid and the tangent is dashed.
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