What is the tangent of x^4-8x^3+7x^2-4,\at x=1 ?

The tangent of x^4-8x^3+7x^2-4,\at x=1 is y=-6x+2

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tangent of x^4-8x^3+7x^2-4,\at x=1

tangent of

x^{4} - 8 x^{3} + 7 x^{2} - 4, at x = 1

y = - 6 x + 2

Solution steps

Find the tangent point:

(1, - 4)

Find the slope of

y = x^{4} - 8 x^{3} + 7 x^{2} - 4 : \frac{d y}{d x} = 4 x^{3} - 24 x^{2} + 14 x

Compute the slope of

y = x^{4} - 8 x^{3} + 7 x^{2} - 4

x = 1 : m = - 6

Find the line with slope m=

- 6

and passing through

(1, - 4) : y = - 6 x + 2

y = - 6 x + 2

f (x) = 7 e^{e}

\frac{d}{d x} (5 x^{3} + 14 x + 23)

e^{f (x)}

y y^{^{'}} = 9 cos (x)

\frac{d}{d x} (\frac{x}{12} - \frac{12}{x})

What is the tangent of x^4-8x^3+7x^2-4,\at x=1 ?
The tangent of x^4-8x^3+7x^2-4,\at x=1 is y=-6x+2