What is the tangent of f(x)=sqrt(x),(25,5) ?

The tangent of f(x)=sqrt(x),(25,5) is y= 1/10 x+5/2

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tangent of f(x)=sqrt(x),(25,5)

tangent of

f (x) = x, (25, 5)

y = \frac{1}{10} x + \frac{5}{2}

Solution steps

Find the slope of

f (x) = x : \frac{df ( x )}{d x} = \frac{1}{2 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{1}{10}

Find the line with slope m=

\frac{1}{10}

and passing through

(25, 5) : y = \frac{1}{10} x + \frac{5}{2}

y = \frac{1}{10} x + \frac{5}{2}

x \to 1 - lim (2 - x^{2})

x = y^{2} + 1, x - y = 1

\int \frac{1}{28} x d x

\frac{8}{y} - \frac{5}{- y}

\int_{0}^{5} 2 x d x

What is the tangent of f(x)=sqrt(x),(25,5) ?
The tangent of f(x)=sqrt(x),(25,5) is y= 1/10 x+5/2