What is the tangent of f(x)=x^3-2x+2,(2,6) ?

The tangent of f(x)=x^3-2x+2,(2,6) is y=10x-14

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tangent of f(x)=x^3-2x+2,(2,6)

tangent of

f (x) = x^{3} - 2 x + 2, (2, 6)

y = 10 x - 14

Solution steps

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

(2, 6) : m = 10

Find the line with slope m=

10

and passing through

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

y = 10 x - 14

4 t \frac{d y}{d t} - 3 y = t

\frac{d}{d x} (- 2 x sin (2 x) + cos (2 x))

\frac{( s + 1 )}{s ( s ^{2} + s + 1 )}

\frac{d y}{d x} + 1 = 2 y

\int e^{r} d r

What is the tangent of f(x)=x^3-2x+2,(2,6) ?
The tangent of f(x)=x^3-2x+2,(2,6) is y=10x-14