What is the tangent of f(x)=2x^3-x^2+9,\at x=2 ?

The tangent of f(x)=2x^3-x^2+9,\at x=2 is y=20x-19

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tangent of f(x)=2x^3-x^2+9,\at x=2

tangent of

f (x) = 2 x^{3} - x^{2} + 9, at x = 2

y = 20 x - 19

Solution steps

Find the tangent point:

(2, 21)

Find the slope of

f (x) = 2 x^{3} - x^{2} + 9 : \frac{df ( x )}{d x} = 6 x^{2} - 2 x

Compute the slope of

f (x) = 2 x^{3} - x^{2} + 9

x = 2 : m = 20

Find the line with slope m=

20

and passing through

(2, 21) : y = 20 x - 19

y = 20 x - 19

x \to \infty lim (- \frac{π}{2})

x \to 0 lim ((1 - \frac{1}{x})^{x^{2}})

\int \frac{1}{( 49 + x ^{2} ) ^{2}} d x

\frac{\partial}{\partial x} ((\frac{y}{x})^{\frac{1}{z}})

\frac{d}{d x} (r (x) (x)^{2} - x^{2})

What is the tangent of f(x)=2x^3-x^2+9,\at x=2 ?
The tangent of f(x)=2x^3-x^2+9,\at x=2 is y=20x-19