Domain and range of g ( x ) = 3 x 2 + 4 x − 2 for the domain, I find, { x | x ≠ 2 }and for the range y = 3 x 2 + 4 x − 2 and when x = 2 − ϵ, y → 3 ( 2 − ϵ ) 2 + 4 ( 2 − ϵ ) − 2 = K − ϵ → − ∞When x = 2 + ϵ, y → 3 ( 2 + ϵ ) 2 + 4 ( 2 + ϵ ) − 2 = K ϵ → ∞Therefore, ran ( g ( x ) ) = ( − ∞ , ∞ )have I done this right or am I missing something? For instance, I thought about the case when x → ∞, but this looks like y → 1 in this case, nothing remarkable. I don't even know if that is what we say when we have an ∞ 2 ∞ ....Do we just say that is one? Or do we say that is ∞ still?

Question

Domain and range of   g  (  x  )  =            3              x        2            +      4              x      −      2      for the domain, I find,   {  x      |    x  ≠  2  }and for the range  y  =                    3                  x          2                +        4                    x        −        2             and when   x  =  2  −  ϵ,   y  →                    3        (        2        −        ϵ                  )          2                +        4                    (        2        −        ϵ        )        −        2              =            K              −        ϵ              →  −  ∞When   x  =  2  +  ϵ,   y  →                    3        (        2        +        ϵ                  )          2                +        4                    (        2        +        ϵ        )        −        2              =            K      ϵ        →  ∞Therefore,   ran  (  g  (  x  )  )  =  (  −  ∞  ,  ∞  )have I done this right or am I missing something? For instance, I thought about the case when   x  →  ∞, but this looks like   y  →  1 in this case, nothing remarkable. I don&#039;t even know if that is what we say when we have an             ∞      2        ∞  ....Do we just say that is one? Or do we say that is   ∞ still?

Karly Donovan · Accepted Answer

Fix   y  ∈      R  . Then you look for   x such that            3              x        2            +      4              x      −      2        =  y  .Solve this equation: you get  3      x    2    +  4  =  x  y  −  2  y  ⇒  3      x    2    −  y  x  +  2  y  +  4  =  0  ,which gives  x  =            y      ±                        y          2                −        24        y        −        48              6    .So your equation admits real solutions only if       y    2    −  24  y  −  48  ≥  0. But this is       y    2    −  24  y  +  144  −  192  =  (  y  −  12      )    2    −  192  ≥  0, i.e.   |  y  −  12  |  ≥      192  . It is clear that this is not true for every   y, since for example   y  =  0 does not satisfy the inequality. So the range is not       R  , but is given by those points   y satisfying   |  y  −  12  |  ≥      192  One more comment on your last observation. If   x  →  ∞ then  g  (  x  )  =                    x        2                    (        3        +                  4                      x            2                          )                    x              (        1        −                  2          x                )              =            x              (        3        +                  4                      x            2                          )                    (      1      −              2        x            )        →            ∞      ⋅      3        1    =  ∞  .

goorst9Bi · Accepted Answer

Your answer is wrong. For example,   0 is not in the range of the function. Here is a hint for finding the range: consider the range on   (  2  ,  ∞  ) and   (  −  ∞  ,  2  ). On these intervals the function is continuous so it will take all values between two limits. So find the max and the min on these two intervals.

Domain and range of g(x) = (3x^2+4)/(x-2)

Answered question

Answer & Explanation

New Questions in Pre-Algebra