Question

find an equation of the line tangent to the graph of $f(x) = \frac{2}{x}$ at $\left(4, \frac{1}{2}\
ight)$

the equation of the tangent line is $y = \square$
(type an expression using $x$ as the variable.)

Explanation:

Step1: Rewrite function for differentiation

$f(x) = 2x^{-1}$

Step2: Compute derivative (slope formula)

$f'(x) = -2x^{-2} = -\frac{2}{x^2}$

Step3: Find slope at $x=4$

$f'(4) = -\frac{2}{4^2} = -\frac{2}{16} = -\frac{1}{8}$

Step4: Use point-slope form

$y - y_1 = m(x - x_1)$ with $(x_1,y_1)=(4,\frac{1}{2}), m=-\frac{1}{8}$
$y - \frac{1}{2} = -\frac{1}{8}(x - 4)$

Step5: Simplify to slope-intercept form

$y = -\frac{1}{8}x + \frac{4}{8} + \frac{1}{2} = -\frac{1}{8}x + \frac{1}{2} + \frac{1}{2} = -\frac{1}{8}x + 1$

Answer:

$-\frac{1}{8}x + 1$