What is the tangent of f(x)= 3/x ,(6, 1/2) ?

The tangent of f(x)= 3/x ,(6, 1/2) is y=-1/12 x+1

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tangent of f(x)= 3/x ,(6, 1/2)

tangent of

f (x) = \frac{3}{x}, (6, \frac{1}{2})

y = - \frac{1}{12} x + 1

Solution steps

Find the slope of

f (x) = \frac{3}{x} : \frac{df ( x )}{d x} = - \frac{3}{x ^{2}}

Compute the slope of

f (x) = \frac{3}{x}

(6, \frac{1}{2}) : m = - \frac{1}{12}

Find the line with slope m=

- \frac{1}{12}

and passing through

(6, \frac{1}{2}) : y = - \frac{1}{12} x + 1

y = - \frac{1}{12} x + 1

\int 32 sin^{4} (2 x) d x

\frac{d}{d y} ((x^{2} + y^{2}) arctan (\frac{y}{x}))

e^{x}, x = 2

\int \frac{x}{7} d x

y^{^{'}} - 5 y = 10 x

What is the tangent of f(x)= 3/x ,(6, 1/2) ?
The tangent of f(x)= 3/x ,(6, 1/2) is y=-1/12 x+1