What is the tangent of f(x)=3+4x^2-2x^3,(2,3) ?

The tangent of f(x)=3+4x^2-2x^3,(2,3) is y=-8x+19

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tangent of f(x)=3+4x^2-2x^3,(2,3)

tangent of

f (x) = 3 + 4 x^{2} - 2 x^{3}, (2, 3)

y = - 8 x + 19

Solution steps

Find the slope of

f (x) = 3 + 4 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 8 x

Compute the slope of

f (x) = 3 + 4 x^{2} - 2 x^{3}

(2, 3) : m = - 8

Find the line with slope m=

- 8

and passing through

(2, 3) : y = - 8 x + 19

y = - 8 x + 19

\int (1 - x^{2})^{3} d x

x \to 0 lim (\frac{9 + x}{x ^{3}})

\frac{\partial}{\partial x} ((2 x^{2} - 5 x)^{3} (2 x + 1))

s \to \infty lim (c t)

What is the tangent of f(x)=3+4x^2-2x^3,(2,3) ?
The tangent of f(x)=3+4x^2-2x^3,(2,3) is y=-8x+19