Is this function increasing/decreasing and convex/concave? y = 3 x + ln ⁡ ( 3 x − 4 x − 1 )

Question

Is this function increasing/decreasing and convex/concave?  y  =  3  x  +  ln  ⁡      (                  3        x        −        4                    x        −        1              )

Reynaldo Hampton · Accepted Answer

First, notice that x is not defined at   x  =  1 and   x  =      4    3  .Set       y    ′    =  0  3  +      1          3              x        2            −      7      x      +      4        =  0  ⇒  3      x    2    −  7  x  +  4  =  −      1    3    ⇒  3      x    2    −  7  x  +      13    3    =  0  (  −  7      )    2    −  4  ∗  3  ∗      13    3    =  −  3  &amp;lt;  0, so it doesn&#039;t have a solution.However, we were assuming   3      x    2    −  7  x  +  4  ≠  0 before. Now consider the case that   3      x    2    −  7  x  +  4  =  0, and we get   x  =  1 or   x  =      4    3  . Neither of the two points fall in the range of   x, so we don&#039;t have to consider these two cases.We can see that       y    ′   is always greater than       0      , which means y is monotonically increasing. And since it is monotonically increasing, it is neither convex nor concave.

Is this function increasing/decreasing and convex/concave? y=3x+ln(3x−4/(x−1))

Answered question

Answer & Explanation

New Questions in Calculus 1