What is the tangent of f(x)=2x^3-4x^2,\at x=1 ?

The tangent of f(x)=2x^3-4x^2,\at x=1 is y=-2x

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tangent of f(x)=2x^3-4x^2,\at x=1

tangent of

f (x) = 2 x^{3} - 4 x^{2}, at x = 1

y = - 2 x

Solution steps

Find the tangent point:

(1, - 2)

Find the slope of

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

Compute the slope of

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

x = 1 : m = - 2

Find the line with slope m=

- 2

and passing through

(1, - 2) : y = - 2 x

y = - 2 x

\int_{1}^{5} \frac{5}{x ^{2} + 4 x + 4} d x

\frac{d}{d x} (x^{2} + 2 x - 3)

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

\frac{d}{d x} (a x e^{b x^{2}})

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

What is the tangent of f(x)=2x^3-4x^2,\at x=1 ?
The tangent of f(x)=2x^3-4x^2,\at x=1 is y=-2x