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

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

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tangent of f(x)=x^3,\at x=-5

tangent of

f (x) = x^{3}, at x = - 5

y = 75 x + 250

Solution steps

Find the tangent point:

(- 5, - 125)

Find the slope of

f (x) = x^{3} : \frac{df ( x )}{d x} = 3 x^{2}

Compute the slope of

f (x) = 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

x \to 0 lim (1 - cos (3 x))

x \to \infty lim (lo g_{2} (lo g_{2} (x) - 1))

\int_{0}^{\frac{π}{2}} sin^{5} (x) d x

\int_{0}^{2} 8000 t e^{- 0.6 t} d t

\int \frac{6 x ^{2} + 3 x + 10}{x ^{3} + 2 x ^{2} + 5 x} d x

What is the tangent of f(x)=x^3,\at x=-5 ?
The tangent of f(x)=x^3,\at x=-5 is y=75x+250