What is the tangent of f(x)=9x-9sqrt(x) ?

The tangent of f(x)=9x-9sqrt(x) is y=(9-9/(2sqrt(a_{0))})x-9sqrt(a_{0)}+(9sqrt(a_{0)})/2

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tangent of f(x)=9x-9sqrt(x)

tangent of

f (x) = 9 x - 9 x

y = (9 - \frac{9}{2 a _{0}}) x - 9 a_{0} + \frac{9 a _{0}}{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 9 a_{0} - 9 a_{0})

Find the slope of

f (x) = 9 x - 9 x : \frac{df ( x )}{d x} = 9 - \frac{9}{2 x}

Compute the slope of

f (x) = 9 x - 9 x

x = a_{0} : m = 9 - \frac{9}{2 a _{0}}

Find the line with slope m=

9 - \frac{9}{2 a _{0}}

and passing through

(a_{0}, 9 a_{0} - 9 a_{0}) : y = (9 - \frac{9}{2 a _{0}}) x - 9 a_{0} + \frac{9 a _{0}}{2}

y = (9 - \frac{9}{2 a _{0}}) x - 9 a_{0} + \frac{9 a _{0}}{2}

f (x) = - 3 cos (x), at x = \frac{- π}{3}

x \to 0 lim ((1 - x)^{- \frac{1}{x}})

\frac{d}{d x} (ln (\frac{x ^{2} - 3 x + 2}{x ^{2} + 1}))

\frac{- 3 x ^{5} - x ^{4} - x ^{3} + 4 x ^{2}}{x ^{2}}

\int x (x^{3} - \frac{1}{x}) d x

What is the tangent of f(x)=9x-9sqrt(x) ?
The tangent of f(x)=9x-9sqrt(x) is y=(9-9/(2sqrt(a_{0))})x-9sqrt(a_{0)}+(9sqrt(a_{0)})/2