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

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

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

tangent of

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

y = - 2 x + 2

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = x^{4} - 3 x^{2} + 2 : \frac{df ( x )}{d x} = 4 x^{3} - 6 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 = - 2

Find the line with slope m=

- 2

and passing through

(1, 0) : y = - 2 x + 2

y = - 2 x + 2

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