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

The tangent of f(x)=x^4-5x^2-x,\at x=1 is y=-7x+2

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

tangent of

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

y = - 7 x + 2

Solution steps

Find the tangent point:

(1, - 5)

Find the slope of

f (x) = x^{4} - 5 x^{2} - x : \frac{df ( x )}{d x} = 4 x^{3} - 10 x - 1

Compute the slope of

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

x = 1 : m = - 7

Find the line with slope m=

- 7

and passing through

(1, - 5) : y = - 7 x + 2

y = - 7 x + 2

y = \frac{3}{x}, (5, \frac{3}{5})

f (x) = (2 x + 3) (x^{2} + 2 x)

\int_{0}^{5} x e^{- 0.6 x} d x

\frac{d}{d x} (cot (x))

y = \frac{e ^{x}}{x + e ^{x}}

What is the tangent of f(x)=x^4-5x^2-x,\at x=1 ?
The tangent of f(x)=x^4-5x^2-x,\at x=1 is y=-7x+2