What is the tangent of f(x)=8x^2-x^3,\at x=2 ?

The tangent of f(x)=8x^2-x^3,\at x=2 is y=20x-16

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tangent of f(x)=8x^2-x^3,\at x=2

tangent of

f (x) = 8 x^{2} - x^{3}, at x = 2

y = 20 x - 16

Solution steps

Find the tangent point:

(2, 24)

Find the slope of

f (x) = 8 x^{2} - x^{3} : \frac{df ( x )}{d x} = 16 x - 3 x^{2}

Compute the slope of

f (x) = 8 x^{2} - x^{3}

x = 2 : m = 20

Find the line with slope m=

20

and passing through

(2, 24) : y = 20 x - 16

y = 20 x - 16

\frac{\partial}{\partial x} (\frac{y}{2 x})

\frac{\partial}{\partial x} (- 4 x y + x^{4} + y^{4})

y^{^{''}} + 4 y = 0

\int x sin (1 - 2 x) d x

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

What is the tangent of f(x)=8x^2-x^3,\at x=2 ?
The tangent of f(x)=8x^2-x^3,\at x=2 is y=20x-16