What is the tangent of f(x)= 8/(x^2+x+2),\at x=2 ?

The tangent of f(x)= 8/(x^2+x+2),\at x=2 is y=-5/8 x+9/4

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tangent of f(x)= 8/(x^2+x+2),\at x=2

tangent of

f (x) = \frac{8}{x ^{2} + x + 2}, at x = 2

y = - \frac{5}{8} x + \frac{9}{4}

Solution steps

Find the tangent point:

(2, 1)

Find the slope of

f (x) = \frac{8}{x ^{2} + x + 2} : \frac{df ( x )}{d x} = - \frac{8 ( 2 x + 1 )}{( x ^{2} + x + 2 ) ^{2}}

Compute the slope of

f (x) = \frac{8}{x ^{2} + x + 2}

x = 2 : m = - \frac{5}{8}

Find the line with slope m=

- \frac{5}{8}

and passing through

(2, 1) : y = - \frac{5}{8} x + \frac{9}{4}

y = - \frac{5}{8} x + \frac{9}{4}

t^{2} \frac{d y}{d t} + 4 t y - y^{3} = 0

h \to + 0 lim (2 x + h)

\frac{\partial}{\partial x} (ln (x^{3} + y^{5}))

\frac{\partial}{\partial y} ((x + y)^{2})

\int_{0}^{1} 4 x 5^{x} d x

What is the tangent of f(x)= 8/(x^2+x+2),\at x=2 ?
The tangent of f(x)= 8/(x^2+x+2),\at x=2 is y=-5/8 x+9/4