What is the tangent of f(x)=x^3-7x+6,\at x=-1 ?

The tangent of f(x)=x^3-7x+6,\at x=-1 is y=-4x+8

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tangent of f(x)=x^3-7x+6,\at x=-1

tangent of

f (x) = x^{3} - 7 x + 6, at x = - 1

y = - 4 x + 8

Solution steps

Find the tangent point:

(- 1, 12)

Find the slope of

f (x) = x^{3} - 7 x + 6 : \frac{df ( x )}{d x} = 3 x^{2} - 7

Compute the slope of

f (x) = x^{3} - 7 x + 6

x = - 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(- 1, 12) : y = - 4 x + 8

y = - 4 x + 8

\int \frac{1}{( x ^{3} + x )} d x

\int \frac{x - 4}{( x + 2 ) ^{2} + 16} d x

x \to 0 lim (\frac{e ^{x^{3}} - 1}{x})

\frac{\partial}{\partial x} (2 x - 2 x y)

\int \frac{2 sin ( x )}{1} d x

What is the tangent of f(x)=x^3-7x+6,\at x=-1 ?
The tangent of f(x)=x^3-7x+6,\at x=-1 is y=-4x+8