What are the critical points of ` f(x)=(x-1)^(2/3)(x+3)` This is what I have so far. f(x)=(x-1)^(2/3)(x+3) =2/3(x-1)^(-1/3)(x+3) + (x-1)^(2/3)(1) What to do next? I am so lost
This problem must be approached in exactly the way you started it! A critical point is a point at which the derivative does not exist or is equal to zero.
So, let's start exactly the way you did. Let's take the derivative by using the multiplication rule:
`(df(x))/(dx) = (x+3)d/dx(x-1)^(2/3) + (x-1)^(2/3) d/dx (x+3)`
`(df(x))/(dx) = 2/3*(x+3)/(x-1)^(1/3) + (x-1)^(2/3)`
Let's multiply the term on the right by `(x-1)^(1/3)` to get a common denominator:
`(df(x))/(dx) = (2/3x+2+x-1)/(x-1)^(1/3)`
Again, we simplify:
`(df(x))/(dx) = (5/3x+1)/(x-1)^(1/3)`
Our critical points will then be where the numerator or denominator are equal to zero. Setting the numerator to zero and solving:
`0 = 5/3x+1`
`-1 = 5/3x`
`-3/5 = x`
So, one critical point will be at `x = -3/5` . Plugging this value into our original function:
`f(-3/5) =(-8/5)^(2/3)*(12/5) ~~ 3.283`
So, our first critical point is at `(-3/5, 3.283)`.
Our second critical point will be where the denominator of our derivative is equal to zero:
`0 = (x-1)^(1/3)`
`0 = x-1`
`1 = x`
So, our second critical point is where the derivative is undefined at `x = 1`.
To find the actual point, let's plug the value into the original function:
`f(1) = 0^(2/3)*4 = 0`
So, our second critical point is at `(1,0)`.
Because there are no other places at which our derivative is zero or undefined, these two points are our full set of critical points: `(1,0)` and `(-3/5, 3.283)`.
Just to confirm, let's graph the function:
Looks like we're right-on! I hope this helps!