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

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

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

tangent of

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

y = 10 x + 10

Solution steps

Find the tangent point:

(- 1, 0)

Find the slope of

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

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 = 10

Find the line with slope m=

10

and passing through

(- 1, 0) : y = 10 x + 10

y = 10 x + 10

\frac{d}{d x} (cos^{n - 1} (x))

y = x^{4} - 4 x^{3} + 10

2 x, x ∣ x + 2 ∣

\frac{d y}{d x} + 3 y = 10, y (0) = 20

x y = 24, 2 x^{2} - y^{2} + 4 = 0

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