What is the tangent of f(x)=sqrt(x)\at (25.5) ?

The tangent of f(x)=sqrt(x)\at (25.5) is y=(12.75}{sqrt(a_{0)}}x+\frac{51sqrt(a_{0)})/4

tangent of f(x)=sqrt(x)\at (25.5)

tangent of

f (x) = x at (25.5)

y = \frac{12.75}{a _{0}} x + \frac{51 a _{0}}{4}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} \cdot 25.5)

Find the slope of

f (x) = x 25.5 : \frac{df ( x )}{d x} = \frac{12.75}{x}

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{12.75}{a _{0}}

Find the line with slope m=

\frac{12.75}{a _{0}}

and passing through

(a_{0}, a_{0} 25.5) : y = \frac{12.75}{a _{0}} x + \frac{51 a _{0}}{4}

y = \frac{12.75}{a _{0}} x + \frac{51 a _{0}}{4}

\int_{1}^{3} (u - 2)^{4} d u

\int e^{8 x} cos (7 x) d x

\int x^{12} d x

x \to 2 - lim (∣ x - 2 ∣)

\frac{\partial}{\partial x} (e^{- x} sin (x + y))

What is the tangent of f(x)=sqrt(x)\at (25.5) ?
The tangent of f(x)=sqrt(x)\at (25.5) is y=(12.75}{sqrt(a_{0)}}x+\frac{51sqrt(a_{0)})/4