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Exercise 28

Chapter P, Section P.1, Page 8
Calculus 9th Edition by Bruce H. Edwards, Ron Larson
ISBN: 9780547167022

Solutions

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Find any intercepts

y=2xx2+1y=2x-\sqrt{x^{2}+1}

Y-intercept

Let call x=0 solve for y

y=2002+1y=01y=01y=1\begin{align*} y&=2\cdot 0 -\sqrt{0^{2}+1}\\ y&=0-\sqrt{1}\\ y&=0-1\\ y&=-1\\ \end{align*}

Y-intercept is at point (0,1)\left( 0,-1\right)

x-intercept

Let y=0 we can solve for x

0=2xx2+1x2+1=2xx2+1=4x23x2+1=0(3x+1)(3x1)=0\begin{align*} 0&=2\cdot x-\sqrt{x^{2}+1}\\ \sqrt{x^{2}+1}&=2\cdot x \\ x^{2}+1&=4\cdot x^{2}\\ -3\cdot x^{2}+1&=0 \\ -\left(\sqrt{3}\cdot x+1 \right)\cdot \left( \sqrt{3}\cdot x -1\right)&=0 \end{align*}

So,

3x+1=03x=1x=13\begin{align*} \sqrt{3}\cdot x+1&=0 \\ \sqrt{3}\cdot x&=-1\\ x&=\dfrac{-1}{\sqrt{3}} \\ \end{align*}

or ,

3x1=03x=1x=13\begin{align*} \sqrt{3}\cdot x-1&=0 \\ \sqrt{3}\cdot x&=1\\ x&=\dfrac{1}{\sqrt{3}} \\ \end{align*}

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