What is the tangent of f(x)=ln(x-4) ?

The tangent of f(x)=ln(x-4) is y= 1/(a_{0)-4}x+ln(a_{0}-4)-(a_{0})/(a_{0)-4}

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tangent of f(x)=ln(x-4)

tangent of

f (x) = ln (x - 4)

y = \frac{1}{a _{0} - 4} x + ln (a_{0} - 4) - \frac{a _{0}}{a _{0} - 4}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, ln (a_{0} - 4))

Find the slope of

f (x) = ln (x - 4) : \frac{df ( x )}{d x} = \frac{1}{x - 4}

Compute the slope of

f (x) = ln (x - 4)

x = a_{0} : m = \frac{1}{a _{0} - 4}

Find the line with slope m=

\frac{1}{a _{0} - 4}

and passing through

(a_{0}, ln (a_{0} - 4)) : y = \frac{1}{a _{0} - 4} x + ln (a_{0} - 4) - \frac{a _{0}}{a _{0} - 4}

y = \frac{1}{a _{0} - 4} x + ln (a_{0} - 4) - \frac{a _{0}}{a _{0} - 4}

\int \frac{x ^{3}}{x ^{2} - 64} d x

f (x) = \frac{1}{x ^{2} + 1}, at x = 2

\frac{d}{d x} (ln (5 e^{- 4 x}))

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

\int \frac{1}{4 - y} d y

What is the tangent of f(x)=ln(x-4) ?
The tangent of f(x)=ln(x-4) is y= 1/(a_{0)-4}x+ln(a_{0}-4)-(a_{0})/(a_{0)-4}