What is the tangent of f(x)=x^4-4x^3+2x^2-6,\at x=1 ?

The tangent of f(x)=x^4-4x^3+2x^2-6,\at x=1 is y=-4x-3

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tangent of f(x)=x^4-4x^3+2x^2-6,\at x=1

tangent of

f (x) = x^{4} - 4 x^{3} + 2 x^{2} - 6, at x = 1

y = - 4 x - 3

Solution steps

Find the tangent point:

(1, - 7)

Find the slope of

f (x) = x^{4} - 4 x^{3} + 2 x^{2} - 6 : \frac{df ( x )}{d x} = 4 x^{3} - 12 x^{2} + 4 x

Compute the slope of

f (x) = x^{4} - 4 x^{3} + 2 x^{2} - 6

x = 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(1, - 7) : y = - 4 x - 3

y = - 4 x - 3

\int \frac{x ^{2}}{x ^{2} + x + 12} d x

\int \frac{1}{- t ^{2} + 3 t - 2} d t

\int e^{\frac{x ^{2}}{4}} d x

x y \frac{d y}{d x} = 7

\int_{0}^{1} sin (π t + \frac{π}{4}) d t

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