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

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

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

tangent of

f (x) = x, 5, 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}

Compute the slope of

f (x) = x

x = 25 : 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 e lim (3 ln (x) - 1)

\int_{1}^{4} \frac{1}{t} d t

\frac{d}{d x} (\frac{2 x ^{3} - 16}{x - 2})

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

f (x) = (x - 6) (2 x + 1)

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