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

The tangent of f(x)=4x^3+3x,\at x=-4,-1 is y=195x+512

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

tangent of

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

y = 195 x + 512

Solution steps

Find the tangent point:

(- 4, - 268)

Find the slope of

f (x) = 4 x^{3} + 3 x : \frac{df ( x )}{d x} = 12 x^{2} + 3

Compute the slope of

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

x = - 4 : m = 195

Find the line with slope m=

195

and passing through

(- 4, - 268) : y = 195 x + 512

y = 195 x + 512

y^{^{'}} + \frac{y}{t} = 6 cos (5 t)

\int x^{2} (x^{3} + 8)^{11} d x

x \to 0 lim (2 x^{2} - 3)

\int_{6}^{12} 6 e^{x} d x

y^{^{'}} = \frac{e ^{t - y}}{1 + e ^{t}}

What is the tangent of f(x)=4x^3+3x,\at x=-4,-1 ?
The tangent of f(x)=4x^3+3x,\at x=-4,-1 is y=195x+512