Could somebody help my find the critical points for y=(x-3)(x+2)^(2/3). When I take the derivative of the function I get is x=-2,3. But when I put the original function in wolfram alpha there is minimum at x=0
your values of x = -2, 3 are the x-intercepts, and you don't need a derivative to find them
dy/dx = (x-3)(2/3)(x+2)^(-1/3) + (x+2)^(2/3)
= (1/3)(x+2)^(-1/3) [2(x-3) + 3(x+2)]
= (1/3)(5x)(x+2)^(-1/3)
= 0 for a max/min of y
----> 5x = 0 or (x+2)^(-1/3) = 0 ---> no way
x = 0, then y = -3(4)^(1/3) or appr -4.76
confirmed by Wolfram
http://www.wolframalpha.com/input/?i=plot+y+%3D+(x-3)(x%2B2)%5E(2%2F3)
To find the critical points of a function, you need to determine where the derivative of the function is equal to zero or undefined.
In this case, to find the critical points of y=(x-3)(x+2)^(2/3), you correctly took the derivative. However, there is one important point to note. When taking the derivative, the exponent rule d/dx (x^n) = nx^(n-1) only applies when the base (x) is positive.
Let's go through the steps to find the critical points:
Step 1: Find the derivative of the function
To find the derivative, you can apply the product rule.
Using the product rule, we have:
dy/dx = (x+2)^(2/3) * d/dx(x-3) + (x-3) * d/dx[(x+2)^(2/3)]
Now, let's calculate the derivatives of each term separately:
d/dx(x-3) = 1 (the derivative of x is 1)
d/dx[(x+2)^(2/3)] = (2/3) * (x+2)^(-1/3) (using the chain rule)
So, the derivative becomes:
dy/dx = (x+2)^(2/3) + (2/3) (x-3) * (x+2)^(-1/3)
Step 2: Set the derivative equal to zero and solve for x
Now, set the derivative equal to zero and solve for x:
(x+2)^(2/3) + (2/3) (x-3) * (x+2)^(-1/3) = 0
To simplify, we can multiply both sides by (x+2)^(1/3) to clear the fraction:
(x+2) + (2/3) (x-3) = 0
3(x+2) + 2(x-3) = 0
3x + 6 + 2x - 6 = 0
5x = 0
x = 0
So, x = 0 is a critical point.
Step 3: Check if the critical point is a maximum or minimum
To determine the nature of the critical point, we need to examine the behavior of the function around x = 0.
Substituting x = 0 back into the original function:
y = (0-3)(0+2)^(2/3)
y = -3(2)^(2/3)
y ≈ -4.732
From this calculation, we can see that the value of y at x = 0 is negative, which implies that there is a local minimum at x = 0. This confirms the result you obtained from Wolfram Alpha.
In summary, the critical points for the function y=(x-3)(x+2)^(2/3) are x = -2, x = 3, and x = 0 (with a local minimum at x = 0).