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

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

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

tangent of

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

y = - 3

Solution steps

Find the slope of

f (x) = \frac{- 6 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = - \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{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 = 0

Find the line with slope m=

0

and passing through

(1, - 3) : y = - 3

y = - 3

\int π x^{4} d x

\frac{x ^{3}}{3} + a x^{2} + b x + 2

y = - x^{2} + 16, y = - \frac{1}{4} x - 1, 4 (y + 2.3) = 2.3 x

\int \frac{22 x ^{2}}{x ^{4} - 61 x ^{2} + 900} d x

x \to 7 lim (\frac{7 - x}{7 - x})

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