What is the tangent of 1/(x+3) ?

The tangent of 1/(x+3) is y=-1/((a_{0)+3)^2}x+(2a_{0}+3)/((a_{0)+3)^2}

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tangent of 1/(x+3)

tangent of

\frac{1}{x + 3}

y = - \frac{1}{( a _{0} + 3 ) ^{2}} x + \frac{2 a _{0} + 3}{( a _{0} + 3 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{a _{0} + 3})

Find the slope of

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

Find the line with slope m=

- \frac{1}{( a _{0} + 3 ) ^{2}}

and passing through

(a_{0}, \frac{1}{a _{0} + 3}) : y = - \frac{1}{( a _{0} + 3 ) ^{2}} x + \frac{2 a _{0} + 3}{( a _{0} + 3 ) ^{2}}

y = - \frac{1}{( a _{0} + 3 ) ^{2}} x + \frac{2 a _{0} + 3}{( a _{0} + 3 ) ^{2}}

What is the tangent of 1/(x+3) ?
The tangent of 1/(x+3) is y=-1/((a_{0)+3)^2}x+(2a_{0}+3)/((a_{0)+3)^2}