What is the tangent of f(x)= 3/(sqrt(x)),\at x= 1/4 ?

The tangent of f(x)= 3/(sqrt(x)),\at x= 1/4 is y=-12x+9

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tangent of f(x)= 3/(sqrt(x)),\at x= 1/4

tangent of

f (x) = \frac{3}{x}, at x = \frac{1}{4}

y = - 12 x + 9

Solution steps

Find the tangent point:

(\frac{1}{4}, 6)

Find the slope of

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

Compute the slope of

f (x) = \frac{3}{x}

x = \frac{1}{4} : m = - 12

Find the line with slope m=

- 12

and passing through

(\frac{1}{4}, 6) : y = - 12 x + 9

y = - 12 x + 9

4 x - 3 y + y^{^{'}} (2 y - 3 x) = 0

t \to 0 lim (\frac{4}{t} - \frac{4}{t ^{2} - t})

\frac{\partial}{\partial x} (- 3 x - 1 e^{x} cos (y))

\int x e^{- x^{2}} d x

\int \frac{y + 1}{y} d y

What is the tangent of f(x)= 3/(sqrt(x)),\at x= 1/4 ?
The tangent of f(x)= 3/(sqrt(x)),\at x= 1/4 is y=-12x+9