What is the tangent of f(x)= 1/x ?

The tangent of f(x)= 1/x is y=-1/(a_{0)^2}x+2/(a_{0)}

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tangent of f(x)= 1/x

tangent of

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

y = - \frac{1}{a _{0}^{2}} x + \frac{2}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{a _{0}})

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{1}{a _{0}^{2}}

Find the line with slope m=

- \frac{1}{a _{0}^{2}}

and passing through

(a_{0}, \frac{1}{a _{0}}) : y = - \frac{1}{a _{0}^{2}} x + \frac{2}{a _{0}}

y = - \frac{1}{a _{0}^{2}} x + \frac{2}{a _{0}}

x \to - 8 lim (\frac{x - 8}{∣ x + 8 ∣})

\int \frac{1}{x ( ln ( x ) ) ^{π}} d x

n = 0 \sum \infty (- 1)^{n} 2^{n}

\int \frac{1}{t ^{2} 121 - t ^{2}} d t

\int e^{2 x} \cdot x^{2} d x

What is the tangent of f(x)= 1/x ?
The tangent of f(x)= 1/x is y=-1/(a_{0)^2}x+2/(a_{0)}