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

The tangent of f(x)=((x+1)/(x-3))^2,\at x=7 is y=-x+11

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tangent of f(x)=((x+1)/(x-3))^2,\at x=7

tangent of

f (x) = (\frac{x + 1}{x - 3})^{2}, at x = 7

y = - x + 11

Solution steps

Find the tangent point:

(7, 4)

Find the slope of

f (x) = (\frac{x + 1}{x - 3})^{2} : \frac{df ( x )}{d x} = - \frac{8 ( x + 1 )}{( x - 3 ) ^{3}}

Compute the slope of

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

x = 7 : m = - 1

Find the line with slope m=

- 1

and passing through

(7, 4) : y = - x + 11

y = - x + 11

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

\frac{\partial}{\partial x} (ln (1 + x y))

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

\frac{\partial}{\partial x} ((e^{x - y}) g oo g)

f (x) = 1 - 2 cos (x)

What is the tangent of f(x)=((x+1)/(x-3))^2,\at x=7 ?
The tangent of f(x)=((x+1)/(x-3))^2,\at x=7 is y=-x+11