What is the tangent of f(x)=x^4-4x^3+8,\at x=2 ?

The tangent of f(x)=x^4-4x^3+8,\at x=2 is y=-16x+24

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tangent of f(x)=x^4-4x^3+8,\at x=2

tangent of

f (x) = x^{4} - 4 x^{3} + 8, at x = 2

y = - 16 x + 24

Solution steps

Find the tangent point:

(2, - 8)

Find the slope of

f (x) = x^{4} - 4 x^{3} + 8 : \frac{df ( x )}{d x} = 4 x^{3} - 12 x^{2}

Compute the slope of

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

x = 2 : m = - 16

Find the line with slope m=

- 16

and passing through

(2, - 8) : y = - 16 x + 24

y = - 16 x + 24

\int 2 x^{3} - 6 x^{2} + 2 x - 1 + (\frac{2}{x}) d x

\frac{\partial}{\partial y} (sin (x y) + cos (x y))

\frac{d}{d x} (\frac{x + 1}{x ^{2} - 1})

\frac{d}{d x} ((\frac{x}{y ^{2}}) - (\frac{y}{x ^{2}}))

y^{^{'}} + 5 y = 4, y (0) = 0

What is the tangent of f(x)=x^4-4x^3+8,\at x=2 ?
The tangent of f(x)=x^4-4x^3+8,\at x=2 is y=-16x+24