What is the tangent of f(x)=sqrt(x-5),\at x=9 ?

The tangent of f(x)=sqrt(x-5),\at x=9 is y= 1/4 x-1/4

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tangent of f(x)=sqrt(x-5),\at x=9

tangent of

f (x) = x - 5, at x = 9

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

Solution steps

Find the tangent point:

(9, 2)

Find the slope of

f (x) = x - 5 : \frac{df ( x )}{d x} = \frac{1}{2 x - 5}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(9, 2) : y = \frac{1}{4} x - \frac{1}{4}

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

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

(x^{2} + 5)^{5}

\int \frac{19}{19 + e ^{x}} d x

\frac{d}{d x} (\frac{1}{3 ^{- x^{2} - y^{2}} + 1})

\frac{d}{d x} (\frac{( x + 1 ) ^{2} ( x - 5 )}{( x - 1 ) ^{3}})

What is the tangent of f(x)=sqrt(x-5),\at x=9 ?
The tangent of f(x)=sqrt(x-5),\at x=9 is y= 1/4 x-1/4