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

The tangent of f(x)=x^3-2x+2,\at x=2 is y=10x-14

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

tangent of

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

y = 10 x - 14

Solution steps

Find the tangent point:

(2, 6)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 10

Find the line with slope m=

10

and passing through

(2, 6) : y = 10 x - 14

y = 10 x - 14

x \to 0 - lim (- 2 x^{2} + 7 x)

y = x^{4} + 6 x^{2} - x, (1, 6)

\frac{\partial}{\partial y} (y^{x} - ln (x^{4} y + 7 x))

\frac{( 2 s + 3 )}{s ^{2} + 2 s + 26}

\int \frac{6 x}{1 - 7 x ^{2}} d x

What is the tangent of f(x)=x^3-2x+2,\at x=2 ?
The tangent of f(x)=x^3-2x+2,\at x=2 is y=10x-14