What is the tangent of f(x)=x^3+2/x ,\at x=1 ?

The tangent of f(x)=x^3+2/x ,\at x=1 is y=x+2

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tangent of f(x)=x^3+2/x ,\at x=1

tangent of

f (x) = x^{3} + \frac{2}{x}, at x = 1

y = x + 2

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = x^{3} + \frac{2}{x} : \frac{df ( x )}{d x} = 3 x^{2} - \frac{2}{x ^{2}}

Compute the slope of

f (x) = x^{3} + \frac{2}{x}

x = 1 : m = 1

Find the line with slope m=

1

and passing through

(1, 3) : y = x + 2

y = x + 2

δ (t)

y^{^{'}} - 2 y = 5

\int \frac{ln ( 6 x )}{x} d x

\int_{6}^{9} (x - 7)^{2} d x

\int cos (x) (7 + 5 sin^{2} (x)) d x

What is the tangent of f(x)=x^3+2/x ,\at x=1 ?
The tangent of f(x)=x^3+2/x ,\at x=1 is y=x+2