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+7

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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 + 7

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 + 7

y = - 3 x + 7

\int 5 e^{- 0.8 x} d x

y = 6 + 4 x csc (x)

n = 1 \sum \infty \frac{3}{( 4 n ^{2} - 1 )}

\frac{\partial}{\partial y} ((2 x + 3 y)^{10})

x y^{^{'}} = y - x^{2} + y^{2}

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+7