What is the tangent of f(x)= 6/(sqrt(x)),\at x=4 ?

The tangent of f(x)= 6/(sqrt(x)),\at x=4 is y=-3/8 x+9/2

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tangent of f(x)= 6/(sqrt(x)),\at x=4

tangent of

f (x) = \frac{6}{x}, at x = 4

y = - \frac{3}{8} x + \frac{9}{2}

Solution steps

Find the tangent point:

(4, 3)

Find the slope of

f (x) = \frac{6}{x} : \frac{df ( x )}{d x} = - \frac{3}{x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{6}{x}

x = 4 : m = - \frac{3}{8}

Find the line with slope m=

- \frac{3}{8}

and passing through

(4, 3) : y = - \frac{3}{8} x + \frac{9}{2}

y = - \frac{3}{8} x + \frac{9}{2}

3 y^{^{'}} + 5 y = 1

\int 2 sin (\frac{t}{2}) d t

4 cos (x)

\frac{\partial}{\partial x} (y - 2 ln ∣ y ∣ + x y)

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

What is the tangent of f(x)= 6/(sqrt(x)),\at x=4 ?
The tangent of f(x)= 6/(sqrt(x)),\at x=4 is y=-3/8 x+9/2