What is the tangent of f(x)=sqrt(x-1),(17,4) ?

The tangent of f(x)=sqrt(x-1),(17,4) is y= 1/8 x+15/8

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tangent of f(x)=sqrt(x-1),(17,4)

tangent of

f (x) = x - 1, (17, 4)

y = \frac{1}{8} x + \frac{15}{8}

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

(17, 4) : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(17, 4) : y = \frac{1}{8} x + \frac{15}{8}

y = \frac{1}{8} x + \frac{15}{8}

n = 0 \sum \infty 4 - sin (n)

\int_{0}^{2} \frac{1}{e ^{π x}} d x

\frac{d y}{d x} + 4 y = e^{- 4 x}

\frac{\partial}{\partial x} (3 x^{2} - 3)

\frac{d}{d x} (a^{2} arcsin (\frac{x}{a}))

What is the tangent of f(x)=sqrt(x-1),(17,4) ?
The tangent of f(x)=sqrt(x-1),(17,4) is y= 1/8 x+15/8