What is the tangent of f(x)=x^5+3x^2+2x,\at x=-1 ?

The tangent of f(x)=x^5+3x^2+2x,\at x=-1 is y=x+1

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tangent of f(x)=x^5+3x^2+2x,\at x=-1

tangent of

f (x) = x^{5} + 3 x^{2} + 2 x, at x = - 1

y = x + 1

Solution steps

Find the tangent point:

(- 1, 0)

Find the slope of

f (x) = x^{5} + 3 x^{2} + 2 x : \frac{df ( x )}{d x} = 5 x^{4} + 6 x + 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 = 1

Find the line with slope m=

1

and passing through

(- 1, 0) : y = x + 1

y = x + 1

3 x^{2} - 4 x + 6

x \to - \infty lim (- 2 e^{x})

\frac{\partial}{\partial y} (\frac{x - y}{x y + 2})

f (5) = 4 x^{2} + 3 x - \frac{8}{x ^{2}}

n = 1 \sum \infty \frac{n}{n ^{2} + 1} x^{n}

What is the tangent of f(x)=x^5+3x^2+2x,\at x=-1 ?
The tangent of f(x)=x^5+3x^2+2x,\at x=-1 is y=x+1