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

The tangent of f(x)=5x^3+4x,\at x=-3 is y=139x+270

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tangent of f(x)=5x^3+4x,\at x=-3

tangent of

f (x) = 5 x^{3} + 4 x, at x = - 3

y = 139 x + 270

Solution steps

Find the tangent point:

(- 3, - 147)

Find the slope of

f (x) = 5 x^{3} + 4 x : \frac{df ( x )}{d x} = 15 x^{2} + 4

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 139

Find the line with slope m=

139

and passing through

(- 3, - 147) : y = 139 x + 270

y = 139 x + 270

f (x) = x \cdot cos (x)

\frac{d}{d x} (((x^{2} + 3)^{5} + x)^{2})

\frac{e ^{x}}{8 + e ^{x}}

\frac{d}{d x} (x + \frac{9}{x})

y = 9 + 3 x + 6 x e^{x}

What is the tangent of f(x)=5x^3+4x,\at x=-3 ?
The tangent of f(x)=5x^3+4x,\at x=-3 is y=139x+270