What is the tangent of f(x)=3+ln(x) ?

The tangent of f(x)=3+ln(x) is y= 1/(a_{0)}x+ln(a_{0})+2

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tangent of f(x)=3+ln(x)

tangent of

f (x) = 3 + ln (x)

y = \frac{1}{a _{0}} x + ln (a_{0}) + 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 + ln (a_{0}))

Find the slope of

f (x) = 3 + ln (x) : \frac{df ( x )}{d x} = \frac{1}{x}

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}{a _{0}}

Find the line with slope m=

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

and passing through

(a_{0}, 3 + ln (a_{0})) : y = \frac{1}{a _{0}} x + ln (a_{0}) + 2

y = \frac{1}{a _{0}} x + ln (a_{0}) + 2

\frac{d}{d x} (- sin (3 x) \cdot 3)

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

\int \frac{3 x + 1}{5 x ^{2} + 1} d x

x \to - 1 lim (- 2)

f (x) = 2 x^{2} + 3 x - 9

What is the tangent of f(x)=3+ln(x) ?
The tangent of f(x)=3+ln(x) is y= 1/(a_{0)}x+ln(a_{0})+2