Question

In Exercises \(1-4,\) find a first-degree polynomial function \(P_{1}\) whose value and slope agree with the value and slope of \(f\) at \(x=c .\) Use a graphing utility to graph \(f\) and \(P_{1} .\) What is \(P_{1}\) called? $$ f(x)=\tan x, \quad c=\frac{\pi}{4} $$

Short answer

The first-degree polynomial function \(P_1\) whose value and slope agree with the value and slope of \(f\) at \(x=c\) is \(P_1 = 2x - \frac{\pi}{2} + 1\). \(P_1\) is called the tangent line to \(f\) at \(x = c\).

Step-by-step solution

Compute \(f(c)\) and \(f'(c)\)

First, we need to find the value and slope of \(f\) at \(x=c=\frac{\pi}{4}\). For the given function \(f(x)=\tan(x)\), it's known that the derivative is \(f'(x)=\sec^2(x)\). So, we compute \(f(c)=\tan(\frac{\pi}{4}) = 1\) and \(f'(c)=\sec^2(\frac{\pi}{4}) = 2\).

Find the equation of the tangent line

We now find the equation of the line that passes through the point \((c,f(c))\) and has slope \(f'(c)\). By the point-slope form of the equation of a line, this line is given by \(y - f(c) = f'(c)(x - c)\), or \(y - 1 = 2(x - \frac{\pi}{4})\).

Rewrite the equation in slope-intercept form

To get \(P_1\), we simplify the equation for \(y\). This gives:\(y = 2x - \frac{\pi}{2} + 1\), so \(P_1 = 2x - \frac{\pi}{2} + 1\).

Identify \(P_1\)

The function \(P_1\) is the first-degree polynomial that we have found. It is called the tangent line to \(f\) at \(x = c\).