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

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

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

tangent of

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

y = - 4 x + 1

Solution steps

Find the tangent point:

(1, - 3)

Find the slope of

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

Compute the slope of

f (x) = x^{4} - 4 x^{2}

x = 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(1, - 3) : y = - 4 x + 1

y = - 4 x + 1

\int 25 - y^{2} d y

3 x - 6, - 2 x + 10, 2, 5

\frac{d y}{d x} + x y = 0

\int \frac{x}{x ^{2} - 5 x + 6} d x

\frac{d}{d x} (1 + x ln (2))

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