If the slope of a line that passes through the origin which is tangent to `y= x^(3) + x + 54` is m, then m is equal to

Answer

To find the slope \( m \) of the line that passes through the origin and is tangent to the curve \( y = x^3 + x + 54 \), we can 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 of the function.

\[ y = x^3 + x + 54 \]

Taking the derivative with respect to \( x \):

\[ \frac{dy}{dx} = 3x^2 + 1 \]

### Step 2: Set up the point of tangency
Let the point of tangency on the curve be \( (x_1, y_1) \). The coordinates of this point can be expressed as:

\[ y_1 = x_1^3 + x_1 + 54 \]

### Step 3: Write the equation of the tangent line
The slope of the tangent line at the point \( (x_1, y_1) \) is given by \( \frac{dy}{dx} \) at \( x_1 \):

\[ m = 3x_1^2 + 1 \]

### Step 4: Use the point-slope form of the line
The equation of the tangent line passing through the origin (0,0) can be written as:

\[ y = mx \]

At the point of tangency, the coordinates must satisfy both the curve and the line:

\[ y_1 = mx_1 \]

Substituting \( y_1 \):

\[ x_1^3 + x_1 + 54 = (3x_1^2 + 1)x_1 \]

### Step 5: Simplify the equation
Expanding the right-hand side:

\[ x_1^3 + x_1 + 54 = 3x_1^3 + x_1 \]

Now, simplify the equation:

\[ x_1^3 + x_1 + 54 - 3x_1^3 - x_1 = 0 \]

This simplifies to:

\[ -2x_1^3 + 54 = 0 \]

### Step 6: Solve for \( x_1 \)
Rearranging gives:

\[ 2x_1^3 = 54 \]

Dividing both sides by 2:

\[ x_1^3 = 27 \]

Taking the cube root:

\[ x_1 = 3 \]

### Step 7: Find the corresponding \( y_1 \)
Now, substitute \( x_1 \) back into the equation for \( y_1 \):

\[ y_1 = 3^3 + 3 + 54 = 27 + 3 + 54 = 84 \]

### Step 8: Calculate the slope \( m \)
Now, substitute \( x_1 = 3 \) into the slope equation:

\[ m = 3(3^2) + 1 = 3(9) + 1 = 27 + 1 = 28 \]

Thus, the slope \( m \) is:

\[ \boxed{28} \]