What is the tangent of f(x)=4+3ln(x),\at x=1 ?

The tangent of f(x)=4+3ln(x),\at x=1 is y=3x+1

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tangent of f(x)=4+3ln(x),\at x=1

tangent of

f (x) = 4 + 3 ln (x), at x = 1

y = 3 x + 1

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = 4 + 3 ln (x) : \frac{df ( x )}{d x} = \frac{3}{x}

Compute the slope of

f (x) = 4 + 3 ln (x)

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, 4) : y = 3 x + 1

y = 3 x + 1

x \to 2 lim (\frac{1}{x})

\int \frac{x ^{3} + x + 1}{x ^{2} - 4 x + 3} d x

\frac{\partial}{\partial z} (x y e^{z} + x - z^{2} - 3)

\int \frac{9}{x ^{2} + 4} d x

\frac{\partial}{\partial x} (6 e^{x} ln (y))

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