What is the tangent of f(x)=x^4+8x^3+2x+2,\at x=1 ?

The tangent of f(x)=x^4+8x^3+2x+2,\at x=1 is y=30x-17

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tangent of f(x)=x^4+8x^3+2x+2,\at x=1

tangent of

f (x) = x^{4} + 8 x^{3} + 2 x + 2, at x = 1

y = 30 x - 17

Solution steps

Find the tangent point:

(1, 13)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 30

Find the line with slope m=

30

and passing through

(1, 13) : y = 30 x - 17

y = 30 x - 17

\frac{\partial}{\partial x} (\frac{12 x}{x + y})

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

\int x^{2} - 2 x + 1 d x

\int x^{2 k} d x

\int \frac{1}{x + 6} d x

What is the tangent of f(x)=x^4+8x^3+2x+2,\at x=1 ?
The tangent of f(x)=x^4+8x^3+2x+2,\at x=1 is y=30x-17