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

The tangent of f(x)= 3/(sqrt(x)),\at x=1 is y=-3/2 x+9/2

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

tangent of

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

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

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

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

Find the line with slope m=

- \frac{3}{2}

and passing through

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

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

f (x) = \frac{3 x ^{2}}{x ^{3} - 8}

4 x^{10 x^{3}}

f (x) = ln (3^{2 - x^{2}})

\int \frac{x + 4}{x + 2} d x

\int (5 x^{3} - 3 x^{2} - 4 x + 50) d x

What is the tangent of f(x)= 3/(sqrt(x)),\at x=1 ?
The tangent of f(x)= 3/(sqrt(x)),\at x=1 is y=-3/2 x+9/2