Hello, I would greatly appreciate your input.
I just need some reassurance that I'm going about this question the right way. I am to find out where h(x)=x*(x-1)^(1/3)is increasing and decreasing. so after I did some solving I found my critical points to be x=(3/4) and x=1.
I tested for some values and stated that h is decreasing at the following intervals: (-∞,3/4) and (3/4,1) and is increasing at (1,∞).
Is this correct? I'm just confused because when I check on a graphing device, it does not show the functions x intervals from (-∞,1), but I thought that this function would have x as a set of real numbers. Am I wrong? If so, please tell me the correct way to go about this.
Thank you for your time. :)
I don't know what you are using to graph your function, but I find this site extremely useful
http://rechneronline.de/function-graphs/
It lets you graph more than one function and it lets you change the ranges for for both x and y
I entered x*(x-1)^(1/3) with the
x-range from -1 to 2 to show your x=3/4 is correct to produce a local minimum.
Experiment with the y-range to get a closer look at the function
:O This graphing device is much better! I was using an application made by google chrome x). Thank you so much, I'll use this from now on :)
Hello! I'd be happy to help you with this question.
You have correctly calculated the critical points of the function h(x) = x*(x-1)^(1/3). The critical points are the values of x where the derivative of h(x) is either zero or undefined. In this case, the derivative is zero at x = 3/4 and x = 1.
To determine where the function is increasing or decreasing, you can analyze the intervals between the critical points.
For x < 3/4: To check if the function is increasing or decreasing in this interval, you can choose a test point x < 3/4 and substitute it into the derivative of h(x). If the derivative is positive, then h(x) is increasing in that interval. If the derivative is negative, then h(x) is decreasing.
For example, let's choose x = 0 as a test point. If we substitute x = 0 into the derivative of h(x), we get h'(0) = (0)*(0-1)^(1/3) = 0. Since the derivative is zero, we cannot determine the sign of the derivative at x = 0 using this test point.
Similarly, for 3/4 < x < 1: You can choose a test point x between 3/4 and 1 and substitute it into the derivative of h(x). If the derivative is positive, then h(x) is increasing in that interval. If the derivative is negative, then h(x) is decreasing.
For x > 1: You can choose a test point x > 1 and substitute it into the derivative. If the derivative is positive, then h(x) is increasing in that interval. If the derivative is negative, then h(x) is decreasing.
Based on your calculations, you have correctly concluded that h(x) is decreasing in the intervals (-∞, 3/4) and (3/4, 1), and increasing in the interval (1, ∞).
Regarding the graphing device not showing the function h(x) for x < 1, it might be due to the fact that the function has a cube root term (x-1)^(1/3) which becomes complex for x < 1. However, mathematically, the function does exist for all real numbers.
I hope this clarifies your doubts! Let me know if you have any further questions.