What is the tangent of f(x)=sqrt(x-1),(5,2) ?

The tangent of f(x)=sqrt(x-1),(5,2) is y= 1/4 x+3/4

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tangent of f(x)=sqrt(x-1),(5,2)

tangent of

f (x) = x - 1, (5, 2)

y = \frac{1}{4} x + \frac{3}{4}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x - 1

(5, 2) : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(5, 2) : y = \frac{1}{4} x + \frac{3}{4}

y = \frac{1}{4} x + \frac{3}{4}

n = 0 \sum \infty \frac{n ^{2}}{3 ^{n}}

e^{x} + 1

\int 9 x^{- 4} d x

\int \frac{3 x + 8}{( x + 4 ) ^{2}} d x

\frac{25}{x ^{2}} - \frac{5}{x}

What is the tangent of f(x)=sqrt(x-1),(5,2) ?
The tangent of f(x)=sqrt(x-1),(5,2) is y= 1/4 x+3/4