Mathematics
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OpenStudy (anonymous):
Is 1 the x-intercept for:
f(x)=x^3-1/x^2-x
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jimthompson5910 (jim_thompson5910):
Hint:
x^3-1 = (x-1)(x^2 + x + 1)
OpenStudy (anonymous):
so there is no x-intercept
jimthompson5910 (jim_thompson5910):
what makes you say that
OpenStudy (anonymous):
x can never = 0
jimthompson5910 (jim_thompson5910):
true, but that doesn't mean we won't have x-intercepts
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OpenStudy (anonymous):
=/
jimthompson5910 (jim_thompson5910):
can you simplify (x^3-1)/(x^2-x)
OpenStudy (anonymous):
would x^2 cancel out and be left with x-1/-x?
jimthompson5910 (jim_thompson5910):
no
jimthompson5910 (jim_thompson5910):
x^3-1 = (x-1)(x^2 + x + 1)
x^2-x = x(x-1)
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jimthompson5910 (jim_thompson5910):
what cancels?
OpenStudy (anonymous):
x+1 and x-1
jimthompson5910 (jim_thompson5910):
you mean the two x-1 terms right?
OpenStudy (anonymous):
correct
jimthompson5910 (jim_thompson5910):
so you're left with
x^2 + x + 1
-----------
x
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jimthompson5910 (jim_thompson5910):
set that whole thing equal to 0 and solve for x
OpenStudy (anonymous):
i see...wouldn't this also mean there would be a Rational discontinuity?
jimthompson5910 (jim_thompson5910):
there is a discontinuity and it's at x = 0 since x can't be 0
OpenStudy (anonymous):
would i have to use the quadratic formula on x^2+x+1 in order to find an x-intercept?
jimthompson5910 (jim_thompson5910):
yes, that's the most direct way to do so (since factoring involves guessing)
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OpenStudy (anonymous):
this would give me an imaginary number though correct?
jimthompson5910 (jim_thompson5910):
that is correct
jimthompson5910 (jim_thompson5910):
so there are no real x intercepts
OpenStudy (anonymous):
awesome! thanks for the help!
jimthompson5910 (jim_thompson5910):
np