What is the tangent of f(x)= 6/(sqrt(x)),(1,6) ?

The tangent of f(x)= 6/(sqrt(x)),(1,6) is y=-3x+9

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tangent of f(x)= 6/(sqrt(x)),(1,6)

tangent of

f (x) = \frac{6}{x}, (1, 6)

y = - 3 x + 9

Solution steps

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}

(1, 6) : m = - 3

Find the line with slope m=

- 3

and passing through

(1, 6) : y = - 3 x + 9

y = - 3 x + 9

\int e^{x + 3 y} d x

\int \frac{sec ^{4} ( x )}{tan ( x )} d x

y = x^{3} - 6 x, (0, 3)

\int \frac{24}{25 + r ^{2}} d r

\int 1 - sin (x) d x

What is the tangent of f(x)= 6/(sqrt(x)),(1,6) ?
The tangent of f(x)= 6/(sqrt(x)),(1,6) is y=-3x+9