What is the tangent of f(x)=x^4-32x^2+256,\at x=-1 ?

The tangent of f(x)=x^4-32x^2+256,\at x=-1 is y=60x+285

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tangent of f(x)=x^4-32x^2+256,\at x=-1

tangent of

f (x) = x^{4} - 32 x^{2} + 256, at x = - 1

y = 60 x + 285

Solution steps

Find the tangent point:

(- 1, 225)

Find the slope of

f (x) = x^{4} - 32 x^{2} + 256 : \frac{df ( x )}{d x} = 4 x^{3} - 64 x

Compute the slope of

f (x) = x^{4} - 32 x^{2} + 256

x = - 1 : m = 60

Find the line with slope m=

60

and passing through

(- 1, 225) : y = 60 x + 285

y = 60 x + 285

\int arctan (7 x) d x

3 e^{t}

y^{^{''}} + 2 y^{^{'}} + y = e^{- t}, y (0) = - 1, y^{^{'}} (0) = 1

\frac{d}{d x} (\frac{60}{x ^{4}})

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

What is the tangent of f(x)=x^4-32x^2+256,\at x=-1 ?
The tangent of f(x)=x^4-32x^2+256,\at x=-1 is y=60x+285