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

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

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tangent of f(x)=sqrt(x),\at x=25

tangent of

f (x) = x, at x = 25

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

Solution steps

Find the tangent point:

(25, 5)

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}

\frac{d}{d x} (- \frac{4}{( x - 4 ) ^{2}})

\int \frac{2 x + 4}{x ^{2} + 4 x + 13} d x

n = 0 \sum \infty 1400700

n = 1 \sum \infty (\frac{5}{n})^{4}

d er i v a t i v e f (x) = \frac{x + 7}{8 x}

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