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

The tangent of f(x)=(x^2-3)(x^3-2x),\at x=-2 is y=26x+48

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

tangent of

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

y = 26 x + 48

Solution steps

Find the tangent point:

(- 2, - 4)

Find the slope of

f (x) = (x^{2} - 3) (x^{3} - 2 x) : \frac{df ( x )}{d x} = 5 x^{4} - 15 x^{2} + 6

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 = 26

Find the line with slope m=

26

and passing through

(- 2, - 4) : y = 26 x + 48

y = 26 x + 48

\frac{\partial}{\partial x} (x e^{y} + x^{2} cos (y))

\int e^{1.25 x} d x

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

f (x) = x^{2} - 1, g (x) = - x + 2, 0, 1

\int (2 - x) 3^{(2 - x)^{2}} d x

What is the tangent of f(x)=(x^2-3)(x^3-2x),\at x=-2 ?
The tangent of f(x)=(x^2-3)(x^3-2x),\at x=-2 is y=26x+48