What is the tangent of f(x)= 1/(sqrt(x)),\at x=1 ?

The tangent of f(x)= 1/(sqrt(x)),\at x=1 is y=-1/2 x+3/2

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tangent of f(x)= 1/(sqrt(x)),\at x=1

tangent of

f (x) = \frac{1}{x}, at x = 1

y = - \frac{1}{2} x + \frac{3}{2}

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

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

Compute the slope of

f (x) = \frac{1}{x}

x = 1 : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(1, 1) : y = - \frac{1}{2} x + \frac{3}{2}

y = - \frac{1}{2} x + \frac{3}{2}

\int e^{- t} cos (t) d t

\frac{d}{d t} (\frac{1}{2} e^{- 4 t} cos (8 t))

\int_{0}^{1} π (x^{4} - x) d x

\frac{\partial}{\partial x} (e^{e^{y} x + y})

f (x) = 4 x^{2} + 6 x + 3

What is the tangent of f(x)= 1/(sqrt(x)),\at x=1 ?
The tangent of f(x)= 1/(sqrt(x)),\at x=1 is y=-1/2 x+3/2