What is the tangent of f(x)=(5x)/(x-3),\at (4) ?

The tangent of f(x)=(5x)/(x-3),\at (4) is y=-(15)/((a_{0)-3)^2}x+(5a_{0}^2)/((a_{0)-3)^2}

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tangent of f(x)=(5x)/(x-3),\at (4)

tangent of

f (x) = \frac{5 x}{x - 3}, at (4)

y = - \frac{15}{( a _{0} - 3 ) ^{2}} x + \frac{5 a _{0}^{2}}{( a _{0} - 3 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{5 a _{0}}{a _{0} - 3})

Find the slope of

f (x) = \frac{5 x}{x - 3} : \frac{df ( x )}{d x} = - \frac{15}{( x - 3 ) ^{2}}

Compute the slope of

f (x) = \frac{5 x}{x - 3}

x = a_{0} : m = - \frac{15}{( a _{0} - 3 ) ^{2}}

Find the line with slope m=

- \frac{15}{( a _{0} - 3 ) ^{2}}

and passing through

(a_{0}, \frac{5 a _{0}}{a _{0} - 3}) : y = - \frac{15}{( a _{0} - 3 ) ^{2}} x + \frac{5 a _{0}^{2}}{( a _{0} - 3 ) ^{2}}

y = - \frac{15}{( a _{0} - 3 ) ^{2}} x + \frac{5 a _{0}^{2}}{( a _{0} - 3 ) ^{2}}

\int \frac{2 + x + x}{x} d x

\int_{0}^{1} x e^{7 x} d x

y = \frac{5}{x} (1.5)

y^{^{''}} + 14 y^{^{'}} + 65 y = 0

f (x) = 3 x + 10, (2, 4)

What is the tangent of f(x)=(5x)/(x-3),\at (4) ?
The tangent of f(x)=(5x)/(x-3),\at (4) is y=-(15)/((a_{0)-3)^2}x+(5a_{0}^2)/((a_{0)-3)^2}