Question

For the function f() = tan¹¹(π) at the point (\left(\frac{1}{4}, 1\right), what is the equation of the tangent line at that point?

Answer 1

The equation of the tangent line to the function
`\(f(x) = \tan^(-1)(\pi)\)` at the point
`\(\left((1)/(4), 1\right)\)` can be found using the point-slope form. The equation is
`\(y - 1 = f'(x_0)(x - (1)/(4))\)`, where
`\(f'(x_0)\)` is the derivative of (f(x)) evaluated at
`\(x = (1)/(4)\).`

Step-by-step explanation:

To determine the equation of the tangent line, we first find the derivative of
`\(f(x) = \tan^(-1)(\pi)\)`. Using the chain rule, the derivative is
`\(f'(x) = (1)/(1 + \pi^2)\)`. Evaluating
`\(f'(x)\) at \(x = (1)/(4)\)`, we get
`\(f'\left((1)/(4)\right) = (1)/(1 + \pi^2)\).`

Now, using the point-slope form of a linear equation
`\(y - y_0 = m(x - x_0)\)`, where (m) is the slope, and
`\((x_0, y_0)\)` is a point on the line, we substitute in the values. The equation of the tangent line is
`\(y - 1 = (1)/(1 + \pi^2)\left(x - (1)/(4)\right)\).`

In conclusion, the equation of the tangent line to
`\(f(x) = \tan^(-1)(\pi)\)` at the point
`\(\left((1)/(4), 1\right)\)` is
`\(y - 1 = (1)/(1 + \pi^2)\left(x - (1)/(4)\right)\)`. This represents a straight line that approximates the behavior of the function near the given point.