What is the tangent of f(x)=9+4x^2-2x^3,\at x=a ?

The tangent of f(x)=9+4x^2-2x^3,\at x=a is y=(-6a^2+8a)x+4a^3-4a^2+9

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tangent of f(x)=9+4x^2-2x^3,\at x=a

tangent of

f (x) = 9 + 4 x^{2} - 2 x^{3}, at x = a

y = (- 6 a^{2} + 8 a) x + 4 a^{3} - 4 a^{2} + 9

Solution steps

Find the tangent point:

(a, 9 + 4 a^{2} - 2 a^{3})

Find the slope of

f (x) = 9 + 4 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 8 x

Compute the slope of

f (x) = 9 + 4 x^{2} - 2 x^{3}

x = a : m = - 6 a^{2} + 8 a

Find the line with slope m=

- 6 a^{2} + 8 a

and passing through

(a, 9 + 4 a^{2} - 2 a^{3}) : y = (- 6 a^{2} + 8 a) x + 4 a^{3} - 4 a^{2} + 9

y = (- 6 a^{2} + 8 a) x + 4 a^{3} - 4 a^{2} + 9

y = 3 x + 4, (7, 5)

y y^{^{'}} = - x

x \to 1 lim (e^{x - 1} - 1)

f (x) = \frac{x ^{2}}{( 4 + x )}

f (x) = - cot (x) csc (x)

What is the tangent of f(x)=9+4x^2-2x^3,\at x=a ?
The tangent of f(x)=9+4x^2-2x^3,\at x=a is y=(-6a^2+8a)x+4a^3-4a^2+9