What is the tangent of f(x)=x-4/x ,\at 0 ?

The tangent of f(x)=x-4/x ,\at 0 is y=(1+4/(a_{0)^2})x-8/(a_{0)}

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tangent of f(x)=x-4/x ,\at 0

tangent of

f (x) = x - \frac{4}{x}, at 0

y = (1 + \frac{4}{a _{0}^{2}}) x - \frac{8}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} - \frac{4}{a _{0}})

Find the slope of

f (x) = x - \frac{4}{x} : \frac{df ( x )}{d x} = 1 + \frac{4}{x ^{2}}

Compute the slope of

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

x = a_{0} : m = 1 + \frac{4}{a _{0}^{2}}

Find the line with slope m=

1 + \frac{4}{a _{0}^{2}}

and passing through

(a_{0}, a_{0} - \frac{4}{a _{0}}) : y = (1 + \frac{4}{a _{0}^{2}}) x - \frac{8}{a _{0}}

y = (1 + \frac{4}{a _{0}^{2}}) x - \frac{8}{a _{0}}

2 y^{^{''}} + 4 y^{^{'}} = 0

y^{^{'}} + 2 y = 2

5 x^{3} lo g_{3} (x)

\int 0.8 sin (x) + \frac{1.6}{π ^{2}} x d x

\frac{\partial}{\partial z} (tan (5 + 2 x^{2} y^{4} z^{2}))

What is the tangent of f(x)=x-4/x ,\at 0 ?
The tangent of f(x)=x-4/x ,\at 0 is y=(1+4/(a_{0)^2})x-8/(a_{0)}