What is the tangent of f(x)=x^9,\at x=-1 ?

The tangent of f(x)=x^9,\at x=-1 is y=9x+8

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tangent of f(x)=x^9,\at x=-1

tangent of

f (x) = x^{9}, at x = - 1

y = 9 x + 8

Solution steps

Find the tangent point:

(- 1, - 1)

Find the slope of

f (x) = x^{9} : \frac{df ( x )}{d x} = 9 x^{8}

Compute the slope of

f (x) = x^{9}

x = - 1 : m = 9

Find the line with slope m=

9

and passing through

(- 1, - 1) : y = 9 x + 8

y = 9 x + 8

\frac{d}{d x} (e^{2 x} n (3 x))

\frac{3}{8} π r^{3}

x \to 0 lim (8 csc (x) - \frac{8}{x})

n = 0 \sum \infty 1^{- n}

\int (3 x) d x

What is the tangent of f(x)=x^9,\at x=-1 ?
The tangent of f(x)=x^9,\at x=-1 is y=9x+8