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

1 Answer

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txmedteach | High School Teacher | (Level 3) Associate Educator

Posted on

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!