What is the tangent of f(x)=5x^3,\at x=-2 ?

The tangent of f(x)=5x^3,\at x=-2 is y=60x+80

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tangent of f(x)=5x^3,\at x=-2

tangent of

f (x) = 5 x^{3}, at x = - 2

y = 60 x + 80

Solution steps

Find the tangent point:

(- 2, - 40)

Find the slope of

f (x) = 5 x^{3} : \frac{df ( x )}{d x} = 15 x^{2}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 60

Find the line with slope m=

60

and passing through

(- 2, - 40) : y = 60 x + 80

y = 60 x + 80

\frac{\partial}{\partial x} (x arcsin (x) y)

\int_{0}^{\frac{π}{2}} x sin (4 x) d x

x \to 5 - lim (\frac{5}{( x - 5 ) ^{3}})

4 x

\int 3 + s (s + 1)^{2} d s

What is the tangent of f(x)=5x^3,\at x=-2 ?
The tangent of f(x)=5x^3,\at x=-2 is y=60x+80