The given function is $y = -\sqrt[5]{2x + 9}$.

To determine where the function is differentiable:

1. Domain restrictions:

   • The expression inside the fifth root must be non-negative because roots of odd degrees (like 5) are defined for all real numbers.
   • However, to avoid undefined situations or complex values for the function, we check: $2x + 9 \geq 0$ Solving this: $x \geq -\frac{9}{2} = -4.5$ So the domain of the function is $x \geq -4.5$.

2. Differentiability:

   • A function is differentiable where it is both continuous and has a well-defined slope (no sharp corners or vertical tangents).
   • For a function involving a root, differentiability is lost at points where the slope becomes vertical or undefined.
   • For $y = -\sqrt[5]{2x + 9}$, differentiability depends on whether the derivative becomes infinite at any point.

The graph shows that the curve behaves smoothly for $x > -4.5$, but at $x = -4.5$, the slope approaches vertical (sharp bend), which indicates the function is not differentiable at $x = -4.5$. It is only differentiable for $x > -4.5$.

Conclusion:

The function is differentiable for $(-4.5, \infty)$.

Would you like further details, or have any other questions? Here are five related questions to consider:

1. What is the derivative of the function $y = -\sqrt[5]{2x + 9}$?
2. How do you determine if a function is differentiable at a specific point?
3. Can you explain the relationship between continuity and differentiability?
4. What happens to the derivative at points where a function has a vertical tangent?
5. How would you graph the derivative of this function?

Tip: Differentiability implies continuity, but continuity does not always imply differentiability. Always check the behavior of the function at key points!