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

The tangent of f(x)= 1/(x^5)-x^2-2x,\at x=1 is y=-9x+7

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

tangent of

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

y = - 9 x + 7

Solution steps

Find the tangent point:

(1, - 2)

Find the slope of

f (x) = \frac{1}{x ^{5}} - x^{2} - 2 x : \frac{df ( x )}{d x} = - \frac{5}{x ^{6}} - 2 x - 2

Compute the slope of

f (x) = \frac{1}{x ^{5}} - x^{2} - 2 x

x = 1 : m = - 9

Find the line with slope m=

- 9

and passing through

(1, - 2) : y = - 9 x + 7

y = - 9 x + 7

f (x) = \frac{x ^{2}}{e ^{x}}

x \to 0 lim (\frac{2}{x} - \frac{2}{x ^{2} + x})

\frac{d}{d x} ((3 x - 4)^{5})

f (t) = sin (t) - t cos (t)

x^{2} y^{^{'}} - 2 x y = 4 y^{4}

What is the tangent of f(x)= 1/(x^5)-x^2-2x,\at x=1 ?
The tangent of f(x)= 1/(x^5)-x^2-2x,\at x=1 is y=-9x+7