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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

(7, 2)

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

x = 7 : 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}

2 t y^{^{'}} = 8 y

\frac{d}{d x} (8 ln (sin (x)))

\int e^{6 x} (6) d x

\frac{d}{d y} (\frac{1}{x y})

\int \frac{6 t ^{3} - t ^{2} + 24 t}{t ^{2} + 4} d t

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