Question

Given $f(x)=2 \sec (x)$, write the equation of the line tangent to $y=f(x)$ when $x=\frac{5 \pi}{3}$.
Answer Attempt 1 out of 2
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Step 1 ：Find the derivative of the function \(f(x) = 2 \sec(x)\).

Step 2 ：The derivative of \(\sec(x)\) is \(\sec(x)\tan(x)\), so the derivative of \(f(x)\) is \(f'(x) = 2\sec(x)\tan(x)\).

Step 3 ：Evaluate the derivative at the point \(x = \frac{5\pi}{3}\) to find the slope of the tangent line at that point.

Step 4 ：\(f'(\frac{5\pi}{3}) = 2\sec(\frac{5\pi}{3})\tan(\frac{5\pi}{3})\).

Step 5 ：We know that \(\sec(\frac{5\pi}{3}) = \frac{1}{\cos(\frac{5\pi}{3})} = -2\) and \(\tan(\frac{5\pi}{3}) = \frac{\sin(\frac{5\pi}{3})}{\cos(\frac{5\pi}{3})} = \sqrt{3}\).

Step 6 ：So, \(f'(\frac{5\pi}{3}) = 2*(-2)*\sqrt{3} = -4\sqrt{3}\).

Step 7 ：This is the slope of the tangent line at the point \(x = \frac{5\pi}{3}\).

Step 8 ：Find the y-coordinate of the point on the function where \(x = \frac{5\pi}{3}\).

Step 9 ：\(f(\frac{5\pi}{3}) = 2\sec(\frac{5\pi}{3}) = 2*(-2) = -4\).

Step 10 ：So, the point on the function where the tangent line touches is \((\frac{5\pi}{3}, -4)\).

Step 11 ：Use the point-slope form of the equation of a line to write the equation of the tangent line. The point-slope form is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope of the line.

Step 12 ：Substituting our values, we get \(y - (-4) = -4\sqrt{3}(x - \frac{5\pi}{3})\).

Step 13 ：Simplifying, we get \(y = -4\sqrt{3}x + 4\sqrt{3}*\frac{5\pi}{3} + 4\).

Step 14 ：So, the equation of the line tangent to \(y = f(x)\) when \(x = \frac{5\pi}{3}\) is \(\boxed{y = -4\sqrt{3}x + 4\sqrt{3}*\frac{5\pi}{3} + 4}\).