SOLUTION: f(x)= 2x^2/ x^4+1 In regards to 27c, d, e, and f. c>> if f(x)=-1, what is x? what points are on the graph of f? My understanding... -1=2x^2/x^4+1 My confusion is real

Question 180779This question is from textbook Sullivan Algebra and Trigonometry
: f(x)= 2x^2/ x^4+1
In regards to 27c, d, e, and f.
c>> if f(x)=-1, what is x? what points are on the graph of f?
My understanding...
-1=2x^2/x^4+1
My confusion is really how to solve for x (due to the fraction with exponents)

d>> what is the domain of f? The answer is all real numbers. Why?
e>> List the X Intercept.
f>> List the Y Intercept.
My understanding...
both intercepts are zero. My graph comes out to be a short line symmertrical to the y axis where both x and y are intercepting at the orgin (0,0) This question is from textbook Sullivan Algebra and Trigonometry

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
f(x)= 2x^2/(x^4+1)
In regards to 27c, d, e, and f.
c>> if f(x)=-1, what is x?
My understanding...
-1=2x^2/x^4+1
Cross-multiply to get:
-x^4-1 = 2x^2
Rearrange:
x^4 + 2x^2 + 1 = 0
Factor:
(x^2+1)(x^2+1) = 0
x = +i or x = -i
-----------------------------
what points are on the graph of f?

-----------------------------------------
f(x) = 2x^2/(x^4+1)
d>> what is the domain of f? The answer is all real numbers. Why?
The domain of a function is always "all Real Numbers" unless the
equation of the function excludes certain values. The x^4+1 in
the denominator is always positive so no x-values are excluded
from the Domain.
-------------------------------------------
e>> List the X Intercept.
f(x) = 2x^2/(x^4+1)
y can only be zero if 2x^2 = 0; so the x-intercept is (0,0)
--------------------------------------------------------------
f>> List the Y Intercept.
If x = 0, y = 0; so the x-intercept is the same as the y-intercept.
================================
Cheers,
Stan H.

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