Question

find an equation of the line tangent to the graph of f(x)=\frac{1}{x^{6}}
a) at (1,1);
b) at (3,\frac{1}{729});
c) at (-2,\frac{1}{64})
a) the equation of the tangent line at (1,1) is y = - 6x + 7 (type an equation using x and y as the variables )
b) the equation of the tangent line at (3,\frac{1}{729}) is (type an equation using x and y as the variables )

Explanation:

Step1: Find derivative of $f(x)$

$f'(x)=-\frac{6}{x^{7}}$

Step2: Evaluate derivative at $x = 3$

$f'(3)=-\frac{6}{3^{7}}=-\frac{6}{2187}=-\frac{2}{729}$

Step3: Use point - slope form $y - y_1=m(x - x_1)$

$y-\frac{1}{729}=-\frac{2}{729}(x - 3)$
$y=-\frac{2}{729}x+\frac{7}{729}$

Answer:

$y=-\frac{2}{729}x+\frac{7}{729}$