What is the tangent of f(x)=x^3-3x+1,(2,3) ?

The tangent of f(x)=x^3-3x+1,(2,3) is y=9x-15

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tangent of f(x)=x^3-3x+1,(2,3)

tangent of

f (x) = x^{3} - 3 x + 1, (2, 3)

y = 9 x - 15

Solution steps

Find the slope of

f (x) = x^{3} - 3 x + 1 : \frac{df ( x )}{d x} = 3 x^{2} - 3

Compute the slope of

f (x) = x^{3} - 3 x + 1

(2, 3) : m = 9

Find the line with slope m=

9

and passing through

(2, 3) : y = 9 x - 15

y = 9 x - 15

\int cos^{3} (5 - x) sin (5 - x) d x

\frac{\partial}{\partial x} (x y - x^{2})

y^{^{'}} = x + y, y (0) = 6

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

\frac{\partial}{\partial s} (s^{2} + t^{2})

What is the tangent of f(x)=x^3-3x+1,(2,3) ?
The tangent of f(x)=x^3-3x+1,(2,3) is y=9x-15