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

Compute the slope of

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

x = - 2 : 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

\frac{21 x ^{2} + 6 x - 6}{2 x x}

\int \frac{1}{9 x ^{2} - 25} d x

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