What is the tangent of y=x^3-3x+1,(4,53) ?

The tangent of y=x^3-3x+1,(4,53) is y=45x-127

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tangent of y=x^3-3x+1,(4,53)

tangent of

y = x^{3} - 3 x + 1, (4, 53)

y = 45 x - 127

Solution steps

Find the slope of

y = x^{3} - 3 x + 1 : \frac{d y}{d x} = 3 x^{2} - 3

Compute the slope of

y = x^{3} - 3 x + 1

(4, 53) : m = 45

Find the line with slope m=

45

and passing through

(4, 53) : y = 45 x - 127

y = 45 x - 127

y = \frac{x}{x + 1}

y^{^{'''}} + 5 y^{^{''}} = 0

\int_{0}^{1} (x - x^{2}) d x

\frac{d}{d x} (- \frac{1}{2 ^{x}})

\frac{\partial}{\partial x} (1 0^{x})

What is the tangent of y=x^3-3x+1,(4,53) ?
The tangent of y=x^3-3x+1,(4,53) is y=45x-127