University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 203: 109


The answer we need to find is $c=4$.

Work Step by Step

$$y=\frac{c}{x+1}$$ 1) Find the derivative of $y$: $$y'=\frac{c'(x+1)-c(x+1)'}{(x+1)^2}=\frac{0(x+1)-c(1)}{(x+1)^2}$$ $$y'=-\frac{c}{(x+1)^2}$$ 2) We need to find the equation of the mentioned line, whose form is following $$y=ax+b$$ - Since the line goes through $(0,3)$: $$a\times0+b=3$$ $$b=3$$ - Since the line goes through $(5,-2)$: $$5a+b=-2$$ Apply $b=3$: $$5a+3=-2$$ $$5a=-5$$ $$a=-1$$ So the mentioned line is $y=-x+3$ 3) A line is tangent to a curve when it intersects the curve at one and only one point, or in the words, the equation $$\frac{c}{x+1}=-x+3$$ has only one solution. $$\frac{c}{x+1}=-x+3$$ $$c=(-x+3)(x+1)=-x^2-x+3x+3$$ $$c=-x^2+2x+3$$ $$x^2-2x+c-3=0$$ Recall that for a quadratic equation $y=Ax^2+Bx+C$, we have $$\Delta=B^2-4AC$$ The equation has only one solution when $\Delta=0$. Here, in this case, $A=1$, $B=-2$ and $C=c-3$ $$\Delta=(-2)^2-4\times1\times(c-3)$$ $$=4-4(c-3)=4-4c+12=16-4c$$ Then, for $\Delta=0$, $$16-4c=0$$ $$c=4$$ So the answer we need to find is $c=4$.
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