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

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

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

tangent of

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

y = - 1

Solution steps

Find the tangent point:

(1, - 1)

Find the slope of

f (x) = x^{4} - 2 x^{2} : \frac{df ( x )}{d x} = 4 x^{3} - 4 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 = 0

Find the line with slope m=

0

and passing through

(1, - 1) : y = - 1

y = - 1

x \to 0 + lim (\frac{1}{x} - cot (x))

\int \frac{5 f ( x )}{x ^{2} ( x + 1 ) ^{3}} d x

\int_{0}^{π} sin (n x) d x

\frac{- cos ( x )}{4 x ^{2} + 2 x - 3}

\frac{\partial}{\partial y} (ln (x + y^{3}))

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