What is the tangent of f(x)=5x^3-5x,(-1,0) ?

The tangent of f(x)=5x^3-5x,(-1,0) is y=10x+10

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tangent of f(x)=5x^3-5x,(-1,0)

tangent of

f (x) = 5 x^{3} - 5 x, (- 1, 0)

y = 10 x + 10

Solution steps

Find the slope of

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

Compute the slope of

f (x) = 5 x^{3} - 5 x

(- 1, 0) : m = 10

Find the line with slope m=

10

and passing through

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

y = 10 x + 10

g (x) = \frac{1 + 2 x}{3 - 4 x}

\int x (4 x + 5)^{3} d x

\frac{d y}{d x}, x y = 28

(- 1.2) (- 2.3)

\int sin^{4} (y) cos (y) d y

What is the tangent of f(x)=5x^3-5x,(-1,0) ?
The tangent of f(x)=5x^3-5x,(-1,0) is y=10x+10