What is the tangent of f(x)=3x^3+5x^2+5x,\at x=-1 ?

The tangent of f(x)=3x^3+5x^2+5x,\at x=-1 is y=4x+1

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tangent of f(x)=3x^3+5x^2+5x,\at x=-1

tangent of

f (x) = 3 x^{3} + 5 x^{2} + 5 x, at x = - 1

y = 4 x + 1

Solution steps

Find the tangent point:

(- 1, - 3)

Find the slope of

f (x) = 3 x^{3} + 5 x^{2} + 5 x : \frac{df ( x )}{d x} = 9 x^{2} + 10 x + 5

Compute the slope of

f (x) = 3 x^{3} + 5 x^{2} + 5 x

x = - 1 : m = 4

Find the line with slope m=

4

and passing through

(- 1, - 3) : y = 4 x + 1

y = 4 x + 1

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

y = 2 x^{2}, y = 2 x + 6, - 2, 2

y \frac{d y}{d x} + 8 x^{3} = 4 x

\int tan (6 x) d x

\frac{d}{d x} (2 e^{x - 5})

What is the tangent of f(x)=3x^3+5x^2+5x,\at x=-1 ?
The tangent of f(x)=3x^3+5x^2+5x,\at x=-1 is y=4x+1