What is the tangent of f(x)=x^3-3x+2,(3,20) ?

The tangent of f(x)=x^3-3x+2,(3,20) is y=24x-52

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tangent of f(x)=x^3-3x+2,(3,20)

tangent of

f (x) = x^{3} - 3 x + 2, (3, 20)

y = 24 x - 52

Solution steps

Find the slope of

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

Compute the slope of

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

(3, 20) : m = 24

Find the line with slope m=

24

and passing through

(3, 20) : y = 24 x - 52

y = 24 x - 52

e^{2 t + 3}

- cos (3 x)

f (x) = \frac{1}{8 x}, at x = 9

\frac{\partial}{\partial x} (2 x^{3} y^{2} + 2 y + 4 x)

\int x^{3} x^{4} + 1 d x

What is the tangent of f(x)=x^3-3x+2,(3,20) ?
The tangent of f(x)=x^3-3x+2,(3,20) is y=24x-52