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

The tangent of f(x)=x^3+3x,\at x=2 is y=15x-16

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

tangent of

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

y = 15 x - 16

Solution steps

Find the tangent point:

(2, 14)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 15

Find the line with slope m=

15

and passing through

(2, 14) : y = 15 x - 16

y = 15 x - 16

\frac{\partial}{\partial x} (x \cdot cos (y))

x \to \infty lim ((x)^{3})

\int \frac{10 x}{1 - x ^{4}} d x

x \to + 0 - lim (\frac{sin ( ∣ x ∣ )}{x})

What is the tangent of f(x)=x^3+3x,\at x=2 ?
The tangent of f(x)=x^3+3x,\at x=2 is y=15x-16