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

The tangent of f(x)=(x^3),\at x=2 is y=12x-16

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

tangent of

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

y = 12 x - 16

Solution steps

Find the tangent point:

(2, 8)

Find the slope of

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

Compute the slope of

f (x) = x^{3}

x = 2 : m = 12

Find the line with slope m=

12

and passing through

(2, 8) : y = 12 x - 16

y = 12 x - 16

x \to 2 lim (\frac{2 - x}{∣ x - 2 ∣})

\frac{d y}{d t} - 2 t y = - 6 t^{2} e^{t^{2}}

\int (cos^{3} (x)) d x

x \to 3 lim (5 x + 2)

x \to 4 lim (32 - x^{2})

What is the tangent of f(x)=(x^3),\at x=2 ?
The tangent of f(x)=(x^3),\at x=2 is y=12x-16