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

The tangent of f(x)=x^2-1/x ,\at x=2 is y= 17/4 x-5

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tangent of f(x)=x^2-1/x ,\at x=2

tangent of

f (x) = x^{2} - \frac{1}{x}, at x = 2

y = \frac{17}{4} x - 5

Solution steps

Find the tangent point:

(2, \frac{7}{2})

Find the slope of

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

Compute the slope of

f (x) = x^{2} - \frac{1}{x}

x = 2 : m = \frac{17}{4}

Find the line with slope m=

\frac{17}{4}

and passing through

(2, \frac{7}{2}) : y = \frac{17}{4} x - 5

y = \frac{17}{4} x - 5

\frac{d}{d y} (e^{\frac{y}{x}})

y^{^{'}} = \frac{( e ^{x} )}{y}

\int (e^{3 x} - 1)^{3} d x

- ln (sin (t))

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

What is the tangent of f(x)=x^2-1/x ,\at x=2 ?
The tangent of f(x)=x^2-1/x ,\at x=2 is y= 17/4 x-5