What is the tangent of f(x)=-2x^5+11x^3,\at x=-2 ?

The tangent of f(x)=-2x^5+11x^3,\at x=-2 is y=-28x-80

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tangent of f(x)=-2x^5+11x^3,\at x=-2

tangent of

f (x) = - 2 x^{5} + 11 x^{3}, at x = - 2

y = - 28 x - 80

Solution steps

Find the tangent point:

(- 2, - 24)

Find the slope of

f (x) = - 2 x^{5} + 11 x^{3} : \frac{df ( x )}{d x} = - 10 x^{4} + 33 x^{2}

Compute the slope of

f (x) = - 2 x^{5} + 11 x^{3}

x = - 2 : m = - 28

Find the line with slope m=

- 28

and passing through

(- 2, - 24) : y = - 28 x - 80

y = - 28 x - 80

y^{^{''}} + 3 y^{^{'}} + y = 0

\int_{2}^{\infty} \frac{1}{e ^{x}} d x

t \to - 2 lim (\frac{t + 2}{t ^{2} - 4})

x \frac{d y}{d x} + 10 y = 9 y^{2}, y (1) = - 1

\frac{d y}{d x} + 4 x^{2} y = - 10 x^{2}

What is the tangent of f(x)=-2x^5+11x^3,\at x=-2 ?
The tangent of f(x)=-2x^5+11x^3,\at x=-2 is y=-28x-80