What is the tangent of f(x)=sqrt(x-3),(7,2) ?

The tangent of f(x)=sqrt(x-3),(7,2) is y= 1/4 x+1/4

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tangent of f(x)=sqrt(x-3),(7,2)

tangent of

f (x) = x - 3, (7, 2)

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

Solution steps

Find the slope of

f (x) = x - 3 : \frac{df ( x )}{d x} = \frac{1}{2 x - 3}

Compute the slope of

f (x) = x - 3

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

Find the line with slope m=

\frac{1}{4}

and passing through

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

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

\frac{d}{d x} (- \frac{6 x}{y})

\frac{9}{x}

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

\int x^{2} 1 0^{x^{3}} d x

n = 0 \sum \infty \frac{n 7 ^{n}}{n !}

What is the tangent of f(x)=sqrt(x-3),(7,2) ?
The tangent of f(x)=sqrt(x-3),(7,2) is y= 1/4 x+1/4