What is the tangent of f(x)= 1/((x-5)^2),\at x=-3 ?

The tangent of f(x)= 1/((x-5)^2),\at x=-3 is y= 1/256 x+7/256

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tangent of f(x)= 1/((x-5)^2),\at x=-3

tangent of

f (x) = \frac{1}{( x - 5 ) ^{2}}, at x = - 3

y = \frac{1}{256} x + \frac{7}{256}

Solution steps

Find the tangent point:

(- 3, \frac{1}{64})

Find the slope of

f (x) = \frac{1}{( x - 5 ) ^{2}} : \frac{df ( x )}{d x} = - \frac{2}{( x - 5 ) ^{3}}

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}{256}

Find the line with slope m=

\frac{1}{256}

and passing through

(- 3, \frac{1}{64}) : y = \frac{1}{256} x + \frac{7}{256}

y = \frac{1}{256} x + \frac{7}{256}

\int - sin (5 x) d x

\frac{ln ( x )}{2 x}

y^{^{''}} + 8 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = 2

\int_{- \infty}^{- 1} \frac{1}{x} d x

\frac{d}{d x} (\frac{x ^{3} + 3}{x})

What is the tangent of f(x)= 1/((x-5)^2),\at x=-3 ?
The tangent of f(x)= 1/((x-5)^2),\at x=-3 is y= 1/256 x+7/256