What is the tangent of f(x)= 7/(sqrt(x)) ?

The tangent of f(x)= 7/(sqrt(x)) is y=-7/(2a_{0)^{3/2}}x+(21)/(2sqrt(a_{0))}

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tangent of f(x)= 7/(sqrt(x))

tangent of

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

y = - \frac{7}{2 a _{0}^{\frac{3}{2}}} x + \frac{21}{2 a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{7}{a _{0}})

Find the slope of

f (x) = \frac{7}{x} : \frac{df ( x )}{d x} = - \frac{7}{2 x ^{\frac{3}{2}}}

Compute the slope of

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

x = a_{0} : m = - \frac{7}{2 a _{0}^{\frac{3}{2}}}

Find the line with slope m=

- \frac{7}{2 a _{0}^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{7}{a _{0}}) : y = - \frac{7}{2 a _{0}^{\frac{3}{2}}} x + \frac{21}{2 a _{0}}

y = - \frac{7}{2 a _{0}^{\frac{3}{2}}} x + \frac{21}{2 a _{0}}

\int \frac{3 x ^{4}}{x ^{2} - 2 x} d x

\frac{- x ^{2} + x + 6}{e ^{x}}

ln (x^{2} - x + 1)

\int \frac{2 x ^{2}}{x ^{4} - 1} d x

y^{^{''}} + \frac{1}{2} y^{^{'}} + \frac{1}{16} y = 0, y (0) = 5, y^{^{'}} (0) = 0

What is the tangent of f(x)= 7/(sqrt(x)) ?
The tangent of f(x)= 7/(sqrt(x)) is y=-7/(2a_{0)^{3/2}}x+(21)/(2sqrt(a_{0))}