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

The tangent of f(x)=x+e^x-1/x ,\at x=1 is y=(e+2)x-2

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

tangent of

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

y = (e + 2) x - 2

Solution steps

Find the tangent point:

(1, e)

Find the slope of

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

Compute the slope of

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

x = 1 : m = e + 2

Find the line with slope m=

e + 2

and passing through

(1, e) : y = (e + 2) x - 2

y = (e + 2) x - 2

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

x \to \frac{π}{3} lim (cos (x))

x \to 1 lim (\frac{- x + 1}{x - 1})

\int 12 x^{2} (2 x^{3} + 3)^{7} d x

\int \frac{x + 5}{x} d x

What is the tangent of f(x)=x+e^x-1/x ,\at x=1 ?
The tangent of f(x)=x+e^x-1/x ,\at x=1 is y=(e+2)x-2