What is the tangent of f(x)=sqrt(x/5),\at x=125 ?

The tangent of f(x)=sqrt(x/5),\at x=125 is y= 1/50 x+5/2

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tangent of f(x)=sqrt(x/5),\at x=125

tangent of

f (x) = \frac{x}{5}, at x = 125

y = \frac{1}{50} x + \frac{5}{2}

Solution steps

Find the tangent point:

(125, 5)

Find the slope of

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

Compute the slope of

f (x) = \frac{x}{5}

x = 125 : m = \frac{1}{50}

Find the line with slope m=

\frac{1}{50}

and passing through

(125, 5) : y = \frac{1}{50} x + \frac{5}{2}

y = \frac{1}{50} x + \frac{5}{2}

\int 3 x e^{- x} d x

\int (y + 1)^{2} d y

x \to 0 lim (\frac{e ^{8 x} - 1}{x})

(2, 1), (- 4, - 2)

\int 4 x^{2} cos (\frac{x}{5}) d x

What is the tangent of f(x)=sqrt(x/5),\at x=125 ?
The tangent of f(x)=sqrt(x/5),\at x=125 is y= 1/50 x+5/2