What is the tangent of f(x)=x^4-8x^3+4,\at x=-1 ?

The tangent of f(x)=x^4-8x^3+4,\at x=-1 is y=-28x-15

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tangent of f(x)=x^4-8x^3+4,\at x=-1

tangent of

f (x) = x^{4} - 8 x^{3} + 4, at x = - 1

y = - 28 x - 15

Solution steps

Find the tangent point:

(- 1, 13)

Find the slope of

f (x) = x^{4} - 8 x^{3} + 4 : \frac{df ( x )}{d x} = 4 x^{3} - 24 x^{2}

Compute the slope of

f (x) = x^{4} - 8 x^{3} + 4

x = - 1 : m = - 28

Find the line with slope m=

- 28

and passing through

(- 1, 13) : y = - 28 x - 15

y = - 28 x - 15

n = 0 \sum \infty \frac{8 n}{6 n + 1}

x y^{^{''}} + 2 y^{^{'}} = 12 x^{2}, y^{^{'}} (1) = 4, y (1) = 0

(x^{2} + \frac{2 y}{x}) d x = (3 - ln (x^{2})) d y

\int 1 + 3 x d x

\int 28 x^{3} - 18 x^{2} + 10 x d x

What is the tangent of f(x)=x^4-8x^3+4,\at x=-1 ?
The tangent of f(x)=x^4-8x^3+4,\at x=-1 is y=-28x-15