y=x^3+px^2+px
find the set of values of p for which this curve has no stationary points

The curve given is \( y = x^3 + px^2 + px \). To find the values of \( p \) such that the curve has no stationary points, we first need to find the derivative of the curve and analyze when it has no stationary points.

### Step 1: Find the derivative
The stationary points occur when the derivative of \( y \) with respect to \( x \), denoted \( \frac{dy}{dx} \), equals zero. Let's differentiate the equation:

\[
\frac{dy}{dx} = 3x^2 + 2px + p
\]

### Step 2: Analyze the derivative
For stationary points to exist, we need to solve \( \frac{dy}{dx} = 0 \), which gives the quadratic equation:

\[
3x^2 + 2px + p = 0
\]

A quadratic equation has no real solutions if its discriminant is negative. The discriminant of the quadratic equation \( ax^2 + bx + c = 0 \) is given by \( \Delta = b^2 - 4ac \). In this case, the discriminant is:

\[
\Delta = (2p)^2 - 4(3)(p) = 4p^2 - 12p
\]

### Step 3: Ensure no stationary points
For the quadratic equation to have no real solutions (i.e., no stationary points), we require the discriminant to be negative:

\[
4p^2 - 12p < 0
\]

### Step 4: Solve the inequality
Factor the expression:

\[
4p(p - 3) < 0
\]

This inequality holds when \( p \) is between 0 and 3. Hence, the curve has no stationary points if:

\[
0 < p < 3
\]

### Final Answer:
The set of values for \( p \) such that the curve has no stationary points is \( p \in (0, 3) \).

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Would you like more details on this process or any other clarifications?

### Related questions:
1. What happens when \( p = 0 \) or \( p = 3 \)?
2. How do we determine the nature of stationary points for \( p \in \mathbb{R} \)?
3. How do stationary points relate to the shape of the curve?
4. What is the significance of the discriminant in quadratic equations?
5. Can this method be applied to higher-degree polynomials?

### Tip:
Whenever solving inequalities involving quadratic expressions, sketching the graph of the quadratic can help visualize where it is positive or negative.

y = x^3 + px^2 + px. Find the set of values of p for which this curve has no stationary points.

Explore how to find the set of values of p such that the curve y = x^3 + px^2 + px has no stationary points. Learn about derivatives, quadratic discriminants, and solving inequalities. Suitable for Grades 11-12 students.

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