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

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

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tangent of f(x)=x-1/x+ln(x),\at x=1

tangent of

f (x) = x - \frac{1}{x} + ln (x), at x = 1

y = 3 x - 3

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = x - \frac{1}{x} + ln (x) : \frac{df ( x )}{d x} = 1 + \frac{1}{x ^{2}} + \frac{1}{x}

Compute the slope of

f (x) = x - \frac{1}{x} + ln (x)

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, 0) : y = 3 x - 3

y = 3 x - 3

y = 16 x sin (x), (\frac{π}{2}, 8 π)

\frac{\partial}{\partial y} (x ln (y + z))

\int \frac{3}{x - 3} d x

x^{2} - 10

ln (x^{4} + 2)

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