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

The tangent of y=(x^4)+(3/((x^2))),\at x=-1 is y=2x+6

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

tangent of

y = (x^{4}) + (\frac{3}{( x ^{2} )}), at x = - 1

y = 2 x + 6

Solution steps

Find the tangent point:

(- 1, 4)

Find the slope of

y = x^{4} + \frac{3}{x ^{2}} : \frac{d y}{d x} = 4 x^{3} - \frac{6}{x ^{3}}

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 = 2

Find the line with slope m=

2

and passing through

(- 1, 4) : y = 2 x + 6

y = 2 x + 6

What is the tangent of y=(x^4)+(3/((x^2))),\at x=-1 ?
The tangent of y=(x^4)+(3/((x^2))),\at x=-1 is y=2x+6