Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises - Page 159: 60

Answer

$(4,3)$

Work Step by Step

The slope of the tangent line to the graph of $y=f(x)$ at $(x,f(x))$ is: $$m_{1}=f'(x)$$ Using the chain rule it follows: $$m_{1}=(\sqrt{1+2x})'=(1+2x)'\cdot \frac{1}{2\sqrt{1+2x}}=2\cdot \frac{1}{2\sqrt{1+2x}}= \frac{1}{\sqrt{1+2x}}$$ The slope intercept form of the line $6x+2y=1$ is $y=-3x+\frac{1}{2}$. So the slope of this line is $m_{2}=-3$. Both lines are perpendicular if the product of their slope is $-1$: $$\frac{1}{\sqrt{1+2x}} \cdot (-3)=-1$$ $$\frac{1}{\sqrt{1+2x}} \cdot 3=1$$ $$ 3=\sqrt{1+2x}$$ $$ 9=1+2x$$ $$9-1=2x$$ $$8=2x$$ $$4=x$$ So the point is $(4,f(4))$ which is $(4,\sqrt{1+2\cdot 4}=3)$ The point is $(4,3)$.
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