What is the tangent of y=x^3-2x+2,(4,58) ?

The tangent of y=x^3-2x+2,(4,58) is y=46x-126

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tangent of y=x^3-2x+2,(4,58)

tangent of

y = x^{3} - 2 x + 2, (4, 58)

y = 46 x - 126

Solution steps

Find the slope of

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

Compute the slope of

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

(4, 58) : m = 46

Find the line with slope m=

46

and passing through

(4, 58) : y = 46 x - 126

y = 46 x - 126

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

\frac{2}{x}, 1

\frac{d y}{d x} = 4 x (x - 1)

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

f (x) = e^{2 x - 5} + 4

What is the tangent of y=x^3-2x+2,(4,58) ?
The tangent of y=x^3-2x+2,(4,58) is y=46x-126