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

Explanation:Given function:f(x)=−1xf′(x)=1x2Now, the slope m of tangent at the given point (3,−13) to the above function:m=f'(3)=132=19Now, the equation of tangent at the point (x1,y1)≡(3,−13) &amp; having slopem=19 is given following formulay−y1=m(x−x1)y−(−13)=19(x−3)9y+3=x-3x-9y-6=0

Explanation:the slope of the tangent line is f'(x) at x=3f(x)=−1x=−x−1f′(x)=x−2=1x2f′(3)=132=19y+13=19(x−3)y+13=19x−13y=19x−103

Explanation:Given function:f(x)=−1xf′(x)=1x2Now, the slope m of tangent at the given point (3,−13) to the above function:m=f'(3)=132=19Now, the equation of tangent at the point (x1,y1)≡(3,−13) &amp; having slopem=19 is given following formulay−y1=m(x−x1)y−(−13)=19(x−3)9y+3=x-3x-9y-6=0

Explanation:the slope of the tangent line is f'(x) at x=3f(x)=−1x=−x−1f′(x)=x−2=1x2f′(3)=132=19y+13=19(x−3)y+13=19x−13y=19x−103

"How do you find the slope & equation..." was answered by our community

ingathat8b9

Answered question

2022-02-13

How do you find the slope & equation of tangent line to

f (x) = - \frac{1}{x}

at (3, -1/3)?

Answer & Explanation

venenolundicoofw

Beginner2022-02-14Added 12 answers

Explanation:
Given function:

f (x) = - \frac{1}{x}

f^{'} (x) = \frac{1}{x^{2}}

Now, the slope m of tangent at the given point

(3, - \frac{1}{3})

to the above function:
m=f'(3)

= \frac{1}{3^{2}}

= \frac{1}{9}

Now, the equation of tangent at the point

(x_{1}, y_{1}) \equiv (3, - \frac{1}{3})

& having slope

m = \frac{1}{9}

is given following formula

y - y_{1} = m (x - x_{1})

y - (- \frac{1}{3}) = \frac{1}{9} (x - 3)

9y+3=x-3
x-9y-6=0

bubble53zjr

Beginner2022-02-15Added 14 answers

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

f (x) = - \frac{1}{x} = - x^{- 1}

f^{'} (x) = x^{- 2} = \frac{1}{x^{2}}

f^{'} (3) = \frac{1}{3^{2}} = \frac{1}{9}

y + \frac{1}{3} = \frac{1}{9} (x - 3)

y + \frac{1}{3} = \frac{1}{9} x - \frac{1}{3}

y = \frac{1}{9} x - \frac{10}{3}

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