What is the tangent of f(x)=x^4-5x^2+4,\at x=-1 ?

The tangent of f(x)=x^4-5x^2+4,\at x=-1 is y=6x+6

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tangent of f(x)=x^4-5x^2+4,\at x=-1

tangent of

f (x) = x^{4} - 5 x^{2} + 4, at x = - 1

y = 6 x + 6

Solution steps

Find the tangent point:

(- 1, 0)

Find the slope of

f (x) = x^{4} - 5 x^{2} + 4 : \frac{df ( x )}{d x} = 4 x^{3} - 10 x

Compute the slope of

f (x) = x^{4} - 5 x^{2} + 4

x = - 1 : m = 6

Find the line with slope m=

6

and passing through

(- 1, 0) : y = 6 x + 6

y = 6 x + 6

x \to 2 lim (cos (\frac{1}{x - 2}))

x \to π lim (3 sin (x))

\int cos (3 x) cos (4 x) d x

\frac{1}{( s + 3 ) ^{2} + 25}

g (x) = (1 - x^{2})^{- x}

What is the tangent of f(x)=x^4-5x^2+4,\at x=-1 ?
The tangent of f(x)=x^4-5x^2+4,\at x=-1 is y=6x+6