Question 1020046
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^5\ -\ x^3].


When *[tex \Large x] is positive, both *[tex \Large x^5] and *[tex \Large x^3] are positive.  When *[tex \Large x] is negative, both *[tex \Large x^5] and *[tex \Large x^3] are negative.  But as *[tex \Large x] gets bigger, *[tex \Large x^5] gets bigger faster than *[tex \Large x^3], so for positive *[tex \Large x], *[tex \Large f(x)] increases without bound as *[tex \Large x] increases without bound. And for negative *[tex \Large x], *[tex \Large f(x)] decreases without bound as *[tex \Large x] decreases without bound.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^3(x^2\ -\ 1)]


Hence if *[tex \Large f(x)\ =\ 0], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ =\ 0\ \Rightarrow\ x\ =\ 0] multiplicity 3


Or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 1\ =\ 0\ \Rightarrow\ (x\ +\ 1)(x\ -\ 1)\ =\ 0]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -1]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 1]


A function crosses the *[tex \Large x] axis whenever *[tex \Large f(x)\ =\ 0].  In your case the function crosses the *[tex \Large x] axis three times, at -1, 0, and 1.


All functions defined at *[tex \Large x\ =\ 0] cross the *[tex \Large y] axis when *[tex \Large x\ =\ 0].  If a function is not defined at *[tex \Large x\ =\ 0], then the function does not cross the *[tex \Large y] axis.


 *[illustration xfifth_xcubed.jpg]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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