What is the tangent of x^3,\at x=-5 ?

The tangent of x^3,\at x=-5 is y=75x+250

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tangent of x^3,\at x=-5

tangent of

x^{3}, at x = - 5

y = 75 x + 250

Solution steps

Find the tangent point:

(- 5, - 125)

Find the slope of

y = x^{3} : \frac{d y}{d x} = 3 x^{2}

Compute the slope of

y = x^{3}

x = - 5 : m = 75

Find the line with slope m=

75

and passing through

(- 5, - 125) : y = 75 x + 250

y = 75 x + 250

\int \frac{1}{- 5 x} d x

\frac{\partial}{\partial x} (4 π^{2} (x^{2} - y^{2}))

\frac{d}{d x} (arcsin (sin (x)))

\int_{0}^{5} \frac{1}{25 + x ^{2}} d x

x \to 0 lim (\frac{x ^{4} + 7 x}{x})