What is the tangent of y=x^3-2x-2,\at x=1 ?

The tangent of y=x^3-2x-2,\at x=1 is y=x-4

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tangent of y=x^3-2x-2,\at x=1

tangent of

y = x^{3} - 2 x - 2, at x = 1

y = x - 4

Solution steps

Find the tangent point:

(1, - 3)

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

x = 1 : m = 1

Find the line with slope m=

1

and passing through

(1, - 3) : y = x - 4

y = x - 4

- sin (x)

\frac{d}{d x} (x^{p} e^{a x})

\int \frac{y}{1 - y ^{2}} d y

\frac{\partial}{\partial x} (- 1 + ln (x))

What is the tangent of y=x^3-2x-2,\at x=1 ?
The tangent of y=x^3-2x-2,\at x=1 is y=x-4