What is the tangent of f(x)=2x^3-x^2+1,\at x=2 ?

The tangent of f(x)=2x^3-x^2+1,\at x=2 is y=20x-27

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tangent of f(x)=2x^3-x^2+1,\at x=2

tangent of

f (x) = 2 x^{3} - x^{2} + 1, at x = 2

y = 20 x - 27

Solution steps

Find the tangent point:

(2, 13)

Find the slope of

f (x) = 2 x^{3} - x^{2} + 1 : \frac{df ( x )}{d x} = 6 x^{2} - 2 x

Compute the slope of

f (x) = 2 x^{3} - x^{2} + 1

x = 2 : m = 20

Find the line with slope m=

20

and passing through

(2, 13) : y = 20 x - 27

y = 20 x - 27

sin (x) - cos (x)

x \frac{d y}{d x} + y = x^{4} ln (x)

x \to 2 + lim (\frac{x + 2}{x ^{2} - 4})

5 + 2 x

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

What is the tangent of f(x)=2x^3-x^2+1,\at x=2 ?
The tangent of f(x)=2x^3-x^2+1,\at x=2 is y=20x-27