Questions: Find the equation of the tangent line to the graph of f at the given point. f(x) = ln(1+2x^8) at (0; 0) The equation of the tangent line to the graph of f at the given point is □ (Use integers or fractions for any numbers in the equation.)

Solution Steps

We start with the function given by

\[ f(x) = \ln(1 + 2x^8). \]

Next, we compute the derivative of the function \( f \):

\[ f'(x) = \frac{d}{dx} \ln(1 + 2x^8) = \frac{16x^7}{1 + 2x^8}. \]

We need to find the slope of the tangent line at the point \( (0, 0) \). We evaluate the derivative at \( x = 0 \):

\[ f'(0) = \frac{16(0)^7}{1 + 2(0)^8} = 0. \]

The slope of the tangent line at the point \( (0, 0) \) is \( m = 0 \). Using the point-slope form of the line equation:

\[ y - y_1 = m(x - x_1), \]

where \( (x_1, y_1) = (0, 0) \), we substitute the values:

\[ y - 0 = 0(x - 0). \]

This simplifies to

\[ y = 0. \]

Final Answer