$$ \begin{array}{l}f(x)=(x-3)^{1 / 3} \ f(x)=|x+3|-\end{array} $$ I need to find the critical points and the open intervals

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To find the intersection points of the two functions, we can set them equal to each other:

$$(x-3)^{1/3} = |x+3|$$

Since the cube root function is always positive, we can remove the absolute value signs:

$$(x-3)^{1/3} = x+3$$

Now, we can cube both sides of the equation to eliminate the cube root:

$$(x-3) = (x+3)^3$$

Expanding the right side:

$$x-3 = (x+3)(x+3)(x+3)$$

$$x-3 = (x+3)(x^2+6x+9)$$

$$x-3 = x^3+9x^2+27x+3x^2+18x+27$$

$$x-3 = x^3+12x^2+45x+27$$

Rearranging the equation to form a cubic equation:

$$x^3+12x^2+45x+27 - x + 3 = 0$$

$$x^3+12x^2+44x+30 = 0$$

Unfortunately, this cubic equation does not have a simple solution. We can use numerical methods or a graphing calculator to find the approximate solutions.The intersection points of the two functions cannot be found algebraically and require numerical methods or a graphing calculator.

\( \begin{array}{l}f(x)=(x-3)^{1 / 3} \\ f(x)=|x+3|-\end{array} \) I need to find the critical points and the open intervals

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