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

The tangent of f(x)=5x^3+4x,\at x=-1 is y=19x+10

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

tangent of

f (x) = 5 x^{3} + 4 x, at x = - 1

y = 19 x + 10

Solution steps

Find the tangent point:

(- 1, - 9)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = 19

Find the line with slope m=

19

and passing through

(- 1, - 9) : y = 19 x + 10

y = 19 x + 10

\frac{d x}{d t} = 1 + t - x - t x

\int 2 (2 x + 4)^{5} d x

\frac{d}{d x} (4^{4})

\int x^{2} cos (\frac{1}{5} x) d x

f (x) = (4 - x) (\frac{2}{x ^{2}} - 3)

What is the tangent of f(x)=5x^3+4x,\at x=-1 ?
The tangent of f(x)=5x^3+4x,\at x=-1 is y=19x+10