What is the tangent of f(x)= 1/x+3,\at x=-1 ?

The tangent of f(x)= 1/x+3,\at x=-1 is y=-x+1

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tangent of f(x)= 1/x+3,\at x=-1

tangent of

f (x) = \frac{1}{x} + 3, at x = - 1

y = - x + 1

Solution steps

Find the tangent point:

(- 1, 2)

Find the slope of

f (x) = \frac{1}{x} + 3 : \frac{df ( x )}{d x} = - \frac{1}{x ^{2}}

Compute the slope of

f (x) = \frac{1}{x} + 3

x = - 1 : m = - 1

Find the line with slope m=

- 1

and passing through

(- 1, 2) : y = - x + 1

y = - x + 1

x \to 5 lim (- \frac{9}{x + 5})

\int_{1}^{3} \frac{1}{3 - x} d x

(sin (6 x))^{4}

\int_{1}^{e} x^{3} ln (x) d x

\frac{d}{d x} (\frac{x}{20})

What is the tangent of f(x)= 1/x+3,\at x=-1 ?
The tangent of f(x)= 1/x+3,\at x=-1 is y=-x+1