# Let `f(x)=x^3+3x^2-24x + 1 ` and` g(x)= x^3 + 6x^2 -36x+1`. For what value(s) of `x` do both `f(x)` and `g(x)` have critical points?So I found derivative of f(x) and g(x) `f(x)=x^3+3x^2-24x + 1`...

Let `f(x)=x^3+3x^2-24x + 1 ` and` g(x)= x^3 + 6x^2 -36x+1`. For what value(s) of `x` do both `f(x)` and `g(x)` have critical points?

So I found derivative of f(x) and g(x)

`f(x)=x^3+3x^2-24x + 1`

`f'(x)=3x^2+6x-24`

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`g(x)= x^3 + 6x^2 -36x+1`

`g'(x)=3x^2+12x-36`

What do I do next?

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You have a great start. When looking at critical points, you are looking for points at which either no derivative exists or the derivative is equal to zero.

With these functions, you very much had the correct idea: you start by taking the derivative.

For f(x):

`f(x) = x^3+3x^2-24x+1`

Taking the derivative:

`(df(x))/(dx) = 3x^2 + 6x - 24`

Similarly for g(x):

`g(x) = x^3 + 6x^2 - 36x + 1`

Take the derivative:

`(dg(x))/(dx) = 3x^2 + 12x - 36`

Now, we have two quadratic functions for our derivatives. Based on our definition for critical points, we simply will set these derivatives to zero. Let's start with `(df)/dx:`

`0 = 3x^2 + 6x - 24`

Now, let's solve this equation like we did during algebra! Let's first simplify by dividing by 3:

`0 = x^2 + 2x - 8`

Now, we can factor this:

`0 = (x+4)(x-2)`

So, the critical points for `f(x)` will be at `x = 2` and `x = -4`.

To find the actual critical points, we simply plug these values of `x` into `f(x)` to find the corresponding y-value. This process gives us the following critical points for f(x):

`(2, -27), (-4, 81)`

Now, for g(x). We do the same thing: set `(dg)/dx` to 0:

`0 = 3x^2 + 12x - 36`

Now, divide by 3 to simplify:

`0 = x^2 + 4x - 12`

We can factor the above equation:

`0 = (x+6)(x-2)`

Giving us `x`-values of -6 and 2 at the critical points. Now, we plug these values of `x` into `g(x)` to determine `y`-values to give us the following critical points.

`(2, -39), (-6, 217)`

I hope that helps!