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Do the following equations define functions: (i) #y = x^2-5x# (ii) #x = y^2-5y# ?

Answer 1

See explanation...

#color(white)()# First equation: #y = x^2-5x#

For the first equation, putting #x=-6# we find:

#y = x^2-5x = (-6)^2-5(-6) = 36+30 = 66#

Note that the value of #y# is uniquely determined by the value of #x#. This is true for any value of #x#, so #y# is a function of #x#.

graph{(y - x^2+5x)(x+6+0.0001y) = 0 [-11, 11, -11, 102]}

#color(white)()# Second equation: #x = y^2-5y#

For the second equation, putting #x=-6# we find:

#-6 = x = y^2-5y#

Adding #6# to both ends we get:

#0 = y^2-5y+6 = (y-2)(y-3)#

So #y = 2# or #y = 3#

Note that #y# is not uniquely determined by the value of #x#, so is not a function of #x#.

graph{(x - y^2+5y)(x+6+0.0001y) = 0 [-12, 2, -2.5, 6]}

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Answer 2

Isaiah Anthony

Yes, both equations define functions.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.