What is the tangent of f(x)=x^4-18x^2+81,\at x=-2 ?

The tangent of f(x)=x^4-18x^2+81,\at x=-2 is y=40x+105

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tangent of f(x)=x^4-18x^2+81,\at x=-2

tangent of

f (x) = x^{4} - 18 x^{2} + 81, at x = - 2

y = 40 x + 105

Solution steps

Find the tangent point:

(- 2, 25)

Find the slope of

f (x) = x^{4} - 18 x^{2} + 81 : \frac{df ( x )}{d x} = 4 x^{3} - 36 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 40

Find the line with slope m=

40

and passing through

(- 2, 25) : y = 40 x + 105

y = 40 x + 105

arcsin (6 x^{4})

x \to \infty lim (\frac{3}{x} + 5 x)

x - 1, y^{2} = 2 x + 6

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

\frac{\partial}{\partial x} (x^{2} (1 + sin (\frac{1}{x})))

What is the tangent of f(x)=x^4-18x^2+81,\at x=-2 ?
The tangent of f(x)=x^4-18x^2+81,\at x=-2 is y=40x+105