What is the tangent of f(x)= 1/(2sqrt(x)),\at x=49 ?

The tangent of f(x)= 1/(2sqrt(x)),\at x=49 is y=-1/1372 x+3/28

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tangent of f(x)= 1/(2sqrt(x)),\at x=49

tangent of

f (x) = \frac{1}{2 x}, at x = 49

y = - \frac{1}{1372} x + \frac{3}{28}

Solution steps

Find the tangent point:

(49, \frac{1}{14})

Find the slope of

f (x) = \frac{1}{2 x} : \frac{df ( x )}{d x} = - \frac{1}{4 x ^{\frac{3}{2}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{1}{1372}

Find the line with slope m=

- \frac{1}{1372}

and passing through

(49, \frac{1}{14}) : y = - \frac{1}{1372} x + \frac{3}{28}

y = - \frac{1}{1372} x + \frac{3}{28}

f (x) = 9 - \frac{4}{x ^{2}}

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

\int x^{3.4} - 2 x^{2 - 1} d x

\int_{- 2}^{0} 4 - x^{2} d x

\int x^{3} sin (x) d x

What is the tangent of f(x)= 1/(2sqrt(x)),\at x=49 ?
The tangent of f(x)= 1/(2sqrt(x)),\at x=49 is y=-1/1372 x+3/28