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

The tangent of f(x)=sqrt(x)+1/x ,\at x=4 is y= 3/16 x+3/2

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

tangent of

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

y = \frac{3}{16} x + \frac{3}{2}

Solution steps

Find the tangent point:

(4, \frac{9}{4})

Find the slope of

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

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{3}{16}

Find the line with slope m=

\frac{3}{16}

and passing through

(4, \frac{9}{4}) : y = \frac{3}{16} x + \frac{3}{2}

y = \frac{3}{16} x + \frac{3}{2}

\int (3 t + 5) cos (\frac{t}{4}) d t

f (x) = 3 x^{(2)} - 5 x - 6

\frac{d}{d x} (16 x)

\frac{d}{d y} (y ln (y) - e^{- x y})

\int \frac{2 sin ( \frac{1}{5 x} )}{5 x ^{2}} d x

What is the tangent of f(x)=sqrt(x)+1/x ,\at x=4 ?
The tangent of f(x)=sqrt(x)+1/x ,\at x=4 is y= 3/16 x+3/2