The number of points on the curve `y=x^(3)-2x^(2)+x-2` where tangents are prarllel to x-axis, is

Options

A. 0
B. 1
C. 2
D. 3

Answer

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

### Step 1: Understand the condition for tangents parallel to the x-axis
A tangent to the curve is parallel to the x-axis when the derivative of the function, \( \frac{dy}{dx} \), is equal to zero.

### Step 2: Differentiate the function
We start with the function:
\[ y = x^3 - 2x^2 + x - 2 \]
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 3x^2 - 4x + 1 \]

### Step 3: Set the derivative equal to zero
To find the points where the tangent is parallel to the x-axis, we set the derivative equal to zero:
\[ 3x^2 - 4x + 1 = 0 \]

### Step 4: Solve the quadratic equation
Now we will solve the quadratic equation \( 3x^2 - 4x + 1 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 3 \), \( b = -4 \), and \( c = 1 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-4)^2 - 4 \cdot 3 \cdot 1 = 16 - 12 = 4 \]
Now substituting into the quadratic formula:
\[ x = \frac{-(-4) \pm \sqrt{4}}{2 \cdot 3} = \frac{4 \pm 2}{6} \]
This gives us two solutions:
1. \( x = \frac{6}{6} = 1 \)
2. \( x = \frac{2}{6} = \frac{1}{3} \)

### Step 5: Count the number of points
We have found two values of \( x \) where the tangent is parallel to the x-axis:
1. \( x = 1 \)
2. \( x = \frac{1}{3} \)

Thus, the number of points on the curve where the tangents are parallel to the x-axis is **2**.

### Final Answer
The number of points on the curve where tangents are parallel to the x-axis is **2**.

---