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

The tangent of f(x)=x^3+2x,\at x=3 is y=29x-54

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

tangent of

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

y = 29 x - 54

Solution steps

Find the tangent point:

(3, 33)

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 = 3 : m = 29

Find the line with slope m=

29

and passing through

(3, 33) : y = 29 x - 54

y = 29 x - 54

x \to 1 lim (7)

x^{4}

v (t) = t - 13, [0, 3]

\int \frac{3 x + 2}{x + 1} d x

2 x^{3} + 10 x^{2} + 22 x - 3, x^{3} - x^{2} - 6 x - 3, - 2, - 2

What is the tangent of f(x)=x^3+2x,\at x=3 ?
The tangent of f(x)=x^3+2x,\at x=3 is y=29x-54