Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter P - P.1 - Graphs and Models - Exercises - Page 9: 76

Answer

a)The intercept with $x$ (or $y$) axis is the point in which the line representing the graph of the function passes through $x$ (or $y$) axis. From the figure given in the problem we see that: For $y=x^3-x$ the intercepts with $x$ axis are points $(-1,0)$ and $(1,0)$ and with $y$ axis the point $(0,0)$. For $y=x^2+2$ the intercepts the graph does not cross the $x$ axis so there are no intercepts with it. The intercept with $y$ axis is at the point $(0,2)$. b) The equation $$y=x^2+2$$ is symmetric with respect to $x$ axis. The equation $$y=x^3-x$$ is symmetric with respect to the origin. c) The point of intersection $(2,6)$.

Work Step by Step

a)The intercept with $x$ (or $y$) axis is the point in which the line representing the graph of the function passes through $x$ (or $y$) axis. From the figure given in the problem we see that: For $y=x^3-x$ the intercepts with $x$ axis are points $(-1,0)$ and $(1,0)$ and with $y$ axis the point $(0,0)$. For $y=x^2+2$ the intercepts the graph does not cross the $x$ axis so there are no intercepts with it. The intercept with $y$ axis is at the point $(0,2)$. b) The equation $$y=x^2+2$$ is symmetric with respect to $x$ axis since is we change $x$ to $-x$ the equation does not change: $$y=(-x)^2+2=x^2+2$$ The equation $$y=x^3-x$$ is symmetric with respect to the origin since it does not change if we put both $-x$ instead of $x$ and $-y$ instead of $y$: $$-y=(-x)^3-(-x)=-x^3+x.$$ Multiplying by $-1$ we obtain the original equation: $$y=x^3-x$$ so it is symmetric with respect to the origin. c) To determine the point of intersection we have to solve the system consisting of the two given equations (find $x$ and $y$ that satisfy both of them): \begin{align} y=x^2+2\\ y=x^3-x\\ \end{align} Since both of the left sides are $y$ we can equate the right sides and calculate $x$: $$x^3-x=x^2+2.$$ Rearranging terms $$x^3-x^2-x-2=0.$$ This is the equation of third degree with whole number coefficients so we will have to guess one solution. Looking at the equation we see that it is satisfied by $x=2$: $$2^3-2^2-2-2=8-4-2-2=0.$$ Now we can factor $$x^3-x^2-x-2=(x-2)(x^2+x+1).$$ The polynomial $x^2+x+1$ has no real zeros (since negative $x$ cannot balance out positive $x^2+1$) so the only solution for $x$ is $x=2$. Now returning this into, lets say, 1st equation we obtain $$y=2^2+2=4+2=6.$$ Now we have for the point of intersection $(2,6)$.
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