What is the tangent of f(x)= 7/(4-x),\at x=-1 ?

The tangent of f(x)= 7/(4-x),\at x=-1 is y= 7/25 x+42/25

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tangent of f(x)= 7/(4-x),\at x=-1

tangent of

f (x) = \frac{7}{4 - x}, at x = - 1

y = \frac{7}{25} x + \frac{42}{25}

Solution steps

Find the tangent point:

(- 1, \frac{7}{5})

Find the slope of

f (x) = \frac{7}{4 - x} : \frac{df ( x )}{d x} = \frac{7}{( 4 - x ) ^{2}}

Compute the slope of

f (x) = \frac{7}{4 - x}

x = - 1 : m = \frac{7}{25}

Find the line with slope m=

\frac{7}{25}

and passing through

(- 1, \frac{7}{5}) : y = \frac{7}{25} x + \frac{42}{25}

y = \frac{7}{25} x + \frac{42}{25}

\frac{\partial}{\partial x} (y^{4} sin (2 x))

x \to - 4 lim (\frac{x ^{2} - 9}{4 - x})

f (x) = 6 e^{- \frac{1}{6} x^{2}}

\int 24 y (1 - x - y) d y

\frac{\partial}{\partial x} (y e^{\frac{x}{y}})

What is the tangent of f(x)= 7/(4-x),\at x=-1 ?
The tangent of f(x)= 7/(4-x),\at x=-1 is y= 7/25 x+42/25