Find the points on the following curve at which the tangents are parallel to x - axis : `y=x^(3)-3x^(2)-9x+7.`

Answer

To find the points on the curve \( y = x^3 - 3x^2 - 9x + 7 \) where the tangents are parallel to the x-axis, we need to follow these steps:

### Step 1: Find the derivative of the function
The slope of the tangent line to the curve at any point is given by the derivative \( \frac{dy}{dx} \). We will differentiate the function \( y \).

\[ y = x^3 - 3x^2 - 9x + 7 \]

Differentiating term by term, we get:

\[ \frac{dy}{dx} = 3x^2 - 6x - 9 \]

### Step 2: Set the derivative equal to zero
For the tangent to be parallel to the x-axis, the slope must be zero. Therefore, we set the derivative equal to zero:

\[ 3x^2 - 6x - 9 = 0 \]

### Step 3: Simplify the equation
We can simplify this equation by dividing all terms by 3:

\[ x^2 - 2x - 3 = 0 \]

### Step 4: Factor the quadratic equation
Next, we will factor the quadratic equation:

\[ (x - 3)(x + 1) = 0 \]

### Step 5: Solve for \( x \)
Setting each factor to zero gives us the values of \( x \):

\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

### Step 6: Find the corresponding \( y \) values
Now, we will substitute these \( x \) values back into the original equation to find the corresponding \( y \) values.

1. For \( x = 3 \):

\[ y = (3)^3 - 3(3)^2 - 9(3) + 7 \]
\[ = 27 - 27 - 27 + 7 = -20 \]

So, the point is \( (3, -20) \).

2. For \( x = -1 \):

\[ y = (-1)^3 - 3(-1)^2 - 9(-1) + 7 \]
\[ = -1 - 3 + 9 + 7 = 12 \]

So, the point is \( (-1, 12) \).

### Final Answer
The points on the curve where the tangents are parallel to the x-axis are:

\[ (3, -20) \quad \text{and} \quad (-1, 12) \]

---