How many tangents are possible from (1, 1) to the curve `y-1=x^3`. Also find these tangents

Answer

To solve the problem of finding how many tangents are possible from the point (1, 1) to the curve \( y - 1 = x^3 \), we can follow these steps:

### Step 1: Rewrite the equation of the curve
The given equation of the curve is:
\[ y - 1 = x^3 \]
This can be rewritten as:
\[ y = x^3 + 1 \]

### Step 2: Find the derivative of the curve
To find the slope of the tangent at any point on the curve, we need to differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 3x^2 \]

### Step 3: Set up the slope condition
Let \( (x_1, y_1) \) be the point on the curve where the tangent touches. The coordinates of this point are:
\[ y_1 = x_1^3 + 1 \]
The slope of the tangent at this point is given by:
\[ m = 3x_1^2 \]

### Step 4: Use the point-slope form of the line
The equation of the tangent line at the point \( (x_1, y_1) \) can be expressed using the point-slope form:
\[ y - y_1 = m(x - x_1) \]
Substituting \( y_1 \) and \( m \):
\[ y - (x_1^3 + 1) = 3x_1^2(x - x_1) \]

### Step 5: Substitute the point (1, 1) into the tangent equation
Since the tangent must pass through the point (1, 1), we substitute \( x = 1 \) and \( y = 1 \):
\[ 1 - (x_1^3 + 1) = 3x_1^2(1 - x_1) \]
Simplifying this gives:
\[ -x_1^3 = 3x_1^2(1 - x_1) \]
\[ -x_1^3 = 3x_1^2 - 3x_1^3 \]
Rearranging terms:
\[ 2x_1^3 - 3x_1^2 = 0 \]

### Step 6: Factor the equation
Factoring out \( x_1^2 \):
\[ x_1^2(2x_1 - 3) = 0 \]
This gives us two solutions:
1. \( x_1^2 = 0 \) which implies \( x_1 = 0 \)
2. \( 2x_1 - 3 = 0 \) which implies \( x_1 = \frac{3}{2} \)

### Step 7: Find the corresponding y-values
For \( x_1 = 0 \):
\[ y_1 = 0^3 + 1 = 1 \]
For \( x_1 = \frac{3}{2} \):
\[ y_1 = \left(\frac{3}{2}\right)^3 + 1 = \frac{27}{8} + 1 = \frac{35}{8} \]

### Step 8: Write the equations of the tangents
1. For \( x_1 = 0 \):
- The slope \( m = 3(0)^2 = 0 \)
- The equation of the tangent is \( y = 1 \).

2. For \( x_1 = \frac{3}{2} \):
- The slope \( m = 3\left(\frac{3}{2}\right)^2 = \frac{27}{4} \)
- Using point-slope form:
\[ y - \frac{35}{8} = \frac{27}{4}\left(x - \frac{3}{2}\right) \]
Rearranging gives:
\[ 4y = 27x - 23 \]

### Conclusion
Thus, there are **2 tangents** possible from the point (1, 1) to the curve \( y - 1 = x^3 \):
1. \( y = 1 \)
2. \( 4y = 27x - 23 \)