What is the tangent of f(x)=(18)/(x^2+2),(1,6) ?

The tangent of f(x)=(18)/(x^2+2),(1,6) is y=-4x+10

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tangent of f(x)=(18)/(x^2+2),(1,6)

tangent of

f (x) = \frac{18}{x ^{2} + 2}, (1, 6)

y = - 4 x + 10

Solution steps

Find the slope of

f (x) = \frac{18}{x ^{2} + 2} : \frac{df ( x )}{d x} = - \frac{36 x}{( x ^{2} + 2 ) ^{2}}

Compute the slope of

f (x) = \frac{18}{x ^{2} + 2}

(1, 6) : m = - 4

Find the line with slope m=

- 4

and passing through

(1, 6) : y = - 4 x + 10

y = - 4 x + 10

x \to 0 + lim (x^{11 x})

\int e^{- 6 x} 9 x d x

\frac{d}{d x} (5 x^{5} sin (2 x))

\frac{d}{d x} (e^{(0.5 x)})

f (x) = ∣ x - 5 ∣

What is the tangent of f(x)=(18)/(x^2+2),(1,6) ?
The tangent of f(x)=(18)/(x^2+2),(1,6) is y=-4x+10