Questions: Find the equation of the tangent line to the graph of f(x)=-1 x^3+3 x^2+4 at x=1.

Solution Steps

To find the equation of the tangent line to the graph of \( f(x) = -x^3 + 3x^2 + 4 \) at \( x = 1 \), we need to:

1. Compute the derivative \( f'(x) \) to find the slope of the tangent line at \( x = 1 \).
2. Evaluate \( f(1) \) to find the y-coordinate of the point of tangency.
3. Use the point-slope form of the equation of a line to write the equation of the tangent line.

To find the slope of the tangent line, we first compute the derivative of the function \( f(x) = -x^3 + 3x^2 + 4 \): \[ f'(x) = -3x^2 + 6x \]

Next, we evaluate the derivative at \( x = 1 \) to find the slope of the tangent line: \[ f'(1) = -3(1)^2 + 6(1) = 3 \]

We also need the y-coordinate of the point of tangency by evaluating the function at \( x = 1 \): \[ f(1) = -1^3 + 3(1)^2 + 4 = 6 \]

Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (1, 6) \) and \( m = 3 \): \[ y - 6 = 3(x - 1) \]

Simplify the equation to get the final form of the tangent line: \[ y - 6 = 3x - 3 \] \[ y = 3x + 3 \]

Final Answer