What is the tangent of f(x)=9x-6sqrt(x),\at x=1 ?

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

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tangent of f(x)=9x-6sqrt(x),\at x=1

tangent of

f (x) = 9 x - 6 x, at x = 1

y = 6 x - 3

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = 9 x - 6 x : \frac{df ( x )}{d x} = 9 - \frac{3}{x}

Compute the slope of

f (x) = 9 x - 6 x

x = 1 : m = 6

Find the line with slope m=

6

and passing through

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

y = 6 x - 3

\frac{d}{d x} ((x)^{4})

\int \frac{sec ^{2} ( \frac{1}{x ^{3}} )}{x ^{4}} d x

\int \frac{cos ( \frac{π}{x ^{19}} )}{x ^{20}} d x

\frac{d y}{d x} = \frac{1}{x} (y^{2} + 0.8 y)

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