Question

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

$f'(-2)=-\frac{6}{(-2)^{7}}=\frac{3}{64}$

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

$y-\frac{1}{64}=\frac{3}{64}(x + 2)$
$y=\frac{3}{64}x+\frac{7}{64}$

Answer:

$y=\frac{3}{64}x+\frac{7}{64}$