What is the tangent of f(x)=5x^3-2x,\at x=-1 ?

The tangent of f(x)=5x^3-2x,\at x=-1 is y=13x+10

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tangent of f(x)=5x^3-2x,\at x=-1

tangent of

f (x) = 5 x^{3} - 2 x, at x = - 1

y = 13 x + 10

Solution steps

Find the tangent point:

(- 1, - 3)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = 13

Find the line with slope m=

13

and passing through

(- 1, - 3) : y = 13 x + 10

y = 13 x + 10

\int x \cdot a c \cdot e^{- a c x + ab c - 1} d x

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

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

y^{^{'}} = x - x^{2}

What is the tangent of f(x)=5x^3-2x,\at x=-1 ?
The tangent of f(x)=5x^3-2x,\at x=-1 is y=13x+10