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

The tangent of f(x)=x^3-2x,\at x=2 is y=10x-16

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

tangent of

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

y = 10 x - 16

Solution steps

Find the tangent point:

(2, 4)

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 = 2 : m = 10

Find the line with slope m=

10

and passing through

(2, 4) : y = 10 x - 16

y = 10 x - 16

\frac{d y}{d x} + 5 (tan (5 x)) y = 3 cos (5 x)

\int 60 x^{2} (1 - 4 x^{3})^{4} d x

f (x) = x^{\frac{5}{6}}

y = 10 - 3 x - x^{2}

What is the tangent of f(x)=x^3-2x,\at x=2 ?
The tangent of f(x)=x^3-2x,\at x=2 is y=10x-16