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

The tangent of f(x)= 1/(sqrt(x)),\at x=4 is y=-1/16 x+3/4

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

tangent of

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

y = - \frac{1}{16} x + \frac{3}{4}

Solution steps

Find the tangent point:

(4, \frac{1}{2})

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 = 4 : m = - \frac{1}{16}

Find the line with slope m=

- \frac{1}{16}

and passing through

(4, \frac{1}{2}) : y = - \frac{1}{16} x + \frac{3}{4}

y = - \frac{1}{16} x + \frac{3}{4}

x \to 5 + lim (\frac{x + 5}{x - 5})

\frac{d}{d x} (sin^{2} (\frac{1}{x}))

\int sin (x) e^{c o s (x)} d x

(4 x^{2} + 9)^{6}

\int_{0}^{1} 5 x^{2} d x

What is the tangent of f(x)= 1/(sqrt(x)),\at x=4 ?
The tangent of f(x)= 1/(sqrt(x)),\at x=4 is y=-1/16 x+3/4