What is the tangent of f(x)=3x^4-4x^2+1,\at x=-1 ?

The tangent of f(x)=3x^4-4x^2+1,\at x=-1 is y=-4x-4

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tangent of f(x)=3x^4-4x^2+1,\at x=-1

tangent of

f (x) = 3 x^{4} - 4 x^{2} + 1, at x = - 1

y = - 4 x - 4

Solution steps

Find the tangent point:

(- 1, 0)

Find the slope of

f (x) = 3 x^{4} - 4 x^{2} + 1 : \frac{df ( x )}{d x} = 12 x^{3} - 8 x

Compute the slope of

f (x) = 3 x^{4} - 4 x^{2} + 1

x = - 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(- 1, 0) : y = - 4 x - 4

y = - 4 x - 4

(x + 1) d y - y d x = 0

x \to 0 lim (\frac{2 x}{cos ( x )})

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

\frac{\partial}{\partial x} (\frac{9}{8 x + 3 y})

\int \frac{x}{1 - 4 x ^{2}} d x

What is the tangent of f(x)=3x^4-4x^2+1,\at x=-1 ?
The tangent of f(x)=3x^4-4x^2+1,\at x=-1 is y=-4x-4