What is the tangent of f(x)=-7x^3-8x^2-5x,\at x=-3 ?

The tangent of f(x)=-7x^3-8x^2-5x,\at x=-3 is y=-146x-306

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tangent of f(x)=-7x^3-8x^2-5x,\at x=-3

tangent of

f (x) = - 7 x^{3} - 8 x^{2} - 5 x, at x = - 3

y = - 146 x - 306

Solution steps

Find the tangent point:

(- 3, 132)

Find the slope of

f (x) = - 7 x^{3} - 8 x^{2} - 5 x : \frac{df ( x )}{d x} = - 21 x^{2} - 16 x - 5

Compute the slope of

f (x) = - 7 x^{3} - 8 x^{2} - 5 x

x = - 3 : m = - 146

Find the line with slope m=

- 146

and passing through

(- 3, 132) : y = - 146 x - 306

y = - 146 x - 306

\int \frac{sin ( 2 x )}{1 + cos ( 2 x )} d x

\int \frac{1}{( 7 x - 1 ) 49 x ^{2} - 14 x} d x

\int_{1}^{2} e^{\frac{1}{x}} d x

\int (\frac{2 y}{( x ^{2} + y ^{2} ) ^{2}}) d y

f (x) = 4 x^{- \frac{3}{8}}

What is the tangent of f(x)=-7x^3-8x^2-5x,\at x=-3 ?
The tangent of f(x)=-7x^3-8x^2-5x,\at x=-3 is y=-146x-306