What is the tangent of f(x)=sqrt(5x-1),\at x=2 ?

The tangent of f(x)=sqrt(5x-1),\at x=2 is y= 5/6 x+4/3

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tangent of f(x)=sqrt(5x-1),\at x=2

tangent of

f (x) = 5 x - 1, at x = 2

y = \frac{5}{6} x + \frac{4}{3}

Solution steps

Find the tangent point:

(2, 3)

Find the slope of

f (x) = 5 x - 1 : \frac{df ( x )}{d x} = \frac{5}{2 5 x - 1}

Compute the slope of

f (x) = 5 x - 1

x = 2 : m = \frac{5}{6}

Find the line with slope m=

\frac{5}{6}

and passing through

(2, 3) : y = \frac{5}{6} x + \frac{4}{3}

y = \frac{5}{6} x + \frac{4}{3}

(1.8) (0.5)

x \to 0 lim (\frac{2}{x ^{3} - x})

y = 2 x^{3} - 13 x, y = x^{2} + 6

x \to 1 - lim (x)

\frac{\partial}{\partial y} ((x + y) \cdot e^{x + 2 y})

What is the tangent of f(x)=sqrt(5x-1),\at x=2 ?
The tangent of f(x)=sqrt(5x-1),\at x=2 is y= 5/6 x+4/3