What is the tangent of f(x)=2x^3-3x^2-16x+3,\at x=-4 ?

The tangent of f(x)=2x^3-3x^2-16x+3,\at x=-4 is y=104x+307

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tangent of f(x)=2x^3-3x^2-16x+3,\at x=-4

tangent of

f (x) = 2 x^{3} - 3 x^{2} - 16 x + 3, at x = - 4

y = 104 x + 307

Solution steps

Find the tangent point:

(- 4, - 109)

Find the slope of

f (x) = 2 x^{3} - 3 x^{2} - 16 x + 3 : \frac{df ( x )}{d x} = 6 x^{2} - 6 x - 16

Compute the slope of

f (x) = 2 x^{3} - 3 x^{2} - 16 x + 3

x = - 4 : m = 104

Find the line with slope m=

104

and passing through

(- 4, - 109) : y = 104 x + 307

y = 104 x + 307

\frac{d}{d x} ((e^{x})^{s i n (x)})

y = x^{3} - 7 x^{2} + 16 x + 7

(0, - 2), (5, - 4)

\int_{0}^{3} 2 e^{- 3 x} d x

\int x^{3} sin (x^{2}) d x

What is the tangent of f(x)=2x^3-3x^2-16x+3,\at x=-4 ?
The tangent of f(x)=2x^3-3x^2-16x+3,\at x=-4 is y=104x+307