What is the tangent of f(x)=2x^3+7x^2-8x-3,\at x=-1 ?

The tangent of f(x)=2x^3+7x^2-8x-3,\at x=-1 is y=-16x-6

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tangent of f(x)=2x^3+7x^2-8x-3,\at x=-1

tangent of

f (x) = 2 x^{3} + 7 x^{2} - 8 x - 3, at x = - 1

y = - 16 x - 6

Solution steps

Find the tangent point:

(- 1, 10)

Find the slope of

f (x) = 2 x^{3} + 7 x^{2} - 8 x - 3 : \frac{df ( x )}{d x} = 6 x^{2} + 14 x - 8

Compute the slope of

f (x) = 2 x^{3} + 7 x^{2} - 8 x - 3

x = - 1 : m = - 16

Find the line with slope m=

- 16

and passing through

(- 1, 10) : y = - 16 x - 6

y = - 16 x - 6

x \to 1 - lim (1 - x - [x] - [1 - x])

x \to 4 - lim (\frac{1}{( x - 4 ) ^{2}})

f (x) = x^{(7)} - 3 e^{(3 x)}

x \to 0 lim ((1 + 4 x)^{\frac{1}{x}})

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 25 e^{- 2 x}

What is the tangent of f(x)=2x^3+7x^2-8x-3,\at x=-1 ?
The tangent of f(x)=2x^3+7x^2-8x-3,\at x=-1 is y=-16x-6