Review the definition of a critical number when finding extrema.menal said:if you plug in zero though, it becomes undefined, not 0...so how can that be a critical point? and do you mean taking (5x-2)/(3x^1/3)
and since the bottom doesn't matter, just find the zero of the top part (5x-2) and get 2/5?
would that be correct?
and the only root?
How did you get 2/5?menal said:oh! lol sorry. i ever knew that! so 0 is a critical number along with 2/5. that would be it?
Right. Now, use the critical numbers to find the extrema.menal said:"you mean taking (5x-2)/(3x^1/3)
and since the bottom doesn't matter, just find the zero of the top part (5x-2) and get 2/5"
The function for finding extrema for (x-1)x^(2/3) is f(x) = (x-1)x^(2/3).
To find the extrema for (x-1)x^(2/3), you can use the first derivative test or the second derivative test. The first derivative test involves finding the critical points of the function and testing them to see if they are local maxima or minima. The second derivative test involves finding the critical points and examining the concavity of the function at those points to determine if they are local maxima or minima.
The domain of (x-1)x^(2/3) is all real numbers except for x = 1, since the function is undefined at that point.
Yes, (x-1)x^(2/3) can have more than one extrema. It can have both a local maximum and a local minimum, or it can have multiple local maxima or minima depending on the shape of the graph.
To graph (x-1)x^(2/3), you can plot a few points and connect them to see the overall shape of the graph. You can also use a graphing calculator or software to plot the function and visualize the extrema. Additionally, you can find the critical points and plug them into the function to determine the exact coordinates of the extrema on the graph.