What is the tangent of f(x)=-x+5x^{-1/2}-3,(1,1) ?

The tangent of f(x)=-x+5x^{-1/2}-3,(1,1) is y=-7/2 x+9/2

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tangent of f(x)=-x+5x^{-1/2}-3,(1,1)

tangent of

f (x) = - x + 5 x^{- \frac{1}{2}} - 3, (1, 1)

y = - \frac{7}{2} x + \frac{9}{2}

Solution steps

Find the slope of

f (x) = - x + 5 x^{- \frac{1}{2}} - 3 : \frac{df ( x )}{d x} = - \frac{5}{2 x ^{\frac{3}{2}}} - 1

Compute the slope of

f (x) = - x + 5 x^{- \frac{1}{2}} - 3

(1, 1) : m = - \frac{7}{2}

Find the line with slope m=

- \frac{7}{2}

and passing through

(1, 1) : y = - \frac{7}{2} x + \frac{9}{2}

y = - \frac{7}{2} x + \frac{9}{2}

\frac{d y}{d x} = \frac{( x - 3 )}{y - 3}

\int_{0}^{\infty} \frac{1}{1 + x} d x

\int 2 cos^{2} (θ) d θ

\frac{d}{d x} (\frac{e ^{x} + e ^{- x}}{e ^{x} - e ^{- x}})

\frac{d ^{3}}{d x ^{3}} (x^{2} sin (x))

What is the tangent of f(x)=-x+5x^{-1/2}-3,(1,1) ?
The tangent of f(x)=-x+5x^{-1/2}-3,(1,1) is y=-7/2 x+9/2