I'm working with critical points and the 1st derivative test and am having a bit of confusion.

The original function

$$f(x) = x(2x-3)^{1/4}$$

has a derivative of

$$f'(x) = \frac{5x-6}{2(2x-3)^{(3/4)}}$$

Setting the numerator to 0 gives me a critical point of 6/5, while a 3/2 in the denominator makes it undefined. But, the graph can't exist at 6/5 = 1.2 because x must be greater than 1.5.

The red line is the original function, the purple curve is the derivative: Graph of this function and its derivative

Furthermore, the derivative never equals 0, so it seems like there can be no critical points except for where f'(x) is undefined, namely, at 3/2. So there is only a single C.P. here.

When I go to make a line graph, all values to the left of 3/2 come up undefined, but all those to the right show than f(x) is increasing. So how can 6/5 be a critical point of the numerator in this case?

Doesn't this mean that there is no place where f(x) is decreasing? Perhaps there is no conflict here but I'm still not sure why the results are so weird. All the examples and practice problems using the line graph have always had either positive or negative values next to the critical point, instead of being undefined to one side.

Line graph with test values greater and less than 3/2:

DNE DNE  ...  ++++++++++++


    0         3/2       2
  • $\begingroup$ Precisely speaking, the function doesn't have any extrema or points of inflection. $\endgroup$ Mar 24, 2015 at 17:47

1 Answer 1


this function does not exist maximum or minimum, it goes up to infinity monotonically as x goes to infinity, and if you take the negative value, it also goes to negative infinity.

  • $\begingroup$ Right, so it's increasing from (1.5, infinity) and has no decreasing values, right? What's up with getting the 6/5 as a possible critical point? $\endgroup$
    – barney
    Mar 24, 2015 at 21:30
  • $\begingroup$ I don't know how did you calculate the derivative, but the derivative that you provided is wrong. $\endgroup$
    – superman
    Mar 24, 2015 at 21:35
  • $\begingroup$ Sorry, your derivative seems to be right, I didn't simplify mine. $\endgroup$
    – superman
    Mar 24, 2015 at 21:41
  • $\begingroup$ The problem is that 6/5 is not on the domain of your derivative function, when you plugin 6/5 to your derivative function, it is undefined.. $\endgroup$
    – superman
    Mar 24, 2015 at 21:44

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