How do you find the slope & equation of tangent line to f(x)=-1/x at (3,-1/3)?

2 Answers

Slope =1/9 & equation: x-9y-6=0

Explanation:

Given function:

f(x)=-1/x

f'(x)=1/x^2

Now, the slope m of tangent at the given point (3, -1/3) to the above function:

m=f'(3)

=1/3^2

=1/9

Now, the equation of tangent at the point (x_1, y_1)\equiv(3, -1/3) & having slope m=1/9 is given following formula

y-y_1=m(x-x_1)

y-(-1/3)=1/9(x-3)

9y+3=x-3

x-9y-6=0

Jul 28, 2018

y=1/9x-10/3

Explanation:

"the slope of the tangent line is "f'(x)" at "x=3

f(x)=-1/x=-x^-1

f'(x)=x^-2=1/x^2

f'(3)=1/(3^2)=1/9

y+1/3=1/9(x-3)

y+1/3=1/9x-1/3

y=1/9x-10/3