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

The tangent of f(x)=x^3+2x,\at x=1 is y=5x-2

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

tangent of

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

y = 5 x - 2

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 5

Find the line with slope m=

5

and passing through

(1, 3) : y = 5 x - 2

y = 5 x - 2

\frac{d}{d x} (\frac{x ^{3} + 3 x + 2}{x ^{2} - 1})

x^{2} y^{^{''}} - 5 x y^{^{'}} + 13 y = 0

\frac{d}{d x} (sin (cos (x^{2} + 3)))

y = 4 x, y = x^{3} + 3 x^{2}

\int_{0}^{1} \frac{2}{7} x^{2} (x + 3) d x

What is the tangent of f(x)=x^3+2x,\at x=1 ?
The tangent of f(x)=x^3+2x,\at x=1 is y=5x-2