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

The tangent of f(x)= 2/(2x+3^2),\at x=-1/3 is y=-36/625 x+138/625

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

tangent of

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

y = - \frac{36}{625} x + \frac{138}{625}

Solution steps

Find the tangent point:

(- \frac{1}{3}, \frac{6}{25})

Find the slope of

f (x) = \frac{2}{2 x + 3 ^{2}} : \frac{df ( x )}{d x} = - \frac{4}{( 2 x + 9 ) ^{2}}

Compute the slope of

f (x) = \frac{2}{2 x + 3 ^{2}}

x = - \frac{1}{3} : m = - \frac{36}{625}

Find the line with slope m=

- \frac{36}{625}

and passing through

(- \frac{1}{3}, \frac{6}{25}) : y = - \frac{36}{625} x + \frac{138}{625}

y = - \frac{36}{625} x + \frac{138}{625}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 6 y = 10 e^{2 x} + 6 x + 7

\frac{d}{d x} (4 x^{2} + 5 x)

\int 4 x^{3} - 20 x^{2} + 8 x + 32 d x

\frac{d}{d x} (- 2 sin (3 x))

f (x) = \frac{5}{x}, at x = 9

What is the tangent of f(x)= 2/(2x+3^2),\at x=-1/3 ?
The tangent of f(x)= 2/(2x+3^2),\at x=-1/3 is y=-36/625 x+138/625