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

The tangent of f(x)=x^5,\at x=-1 is y=5x+4

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

tangent of

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

y = 5 x + 4

Solution steps

Find the tangent point:

(- 1, - 1)

Find the slope of

f (x) = x^{5} : \frac{df ( x )}{d x} = 5 x^{4}

Compute the slope of

f (x) = x^{5}

x = - 1 : m = 5

Find the line with slope m=

5

and passing through

(- 1, - 1) : y = 5 x + 4

y = 5 x + 4

\int x^{2} x^{3} + 39 d x

\int \frac{x}{( 1 - x ) ^{2}} d x

x \to 0 lim (\frac{9 x ^{2} - 8 x}{x})

- y + y^{^{'}} = x + e^{2} x

y^{^{'}} + \frac{2}{x} y = x^{2} y^{2}

What is the tangent of f(x)=x^5,\at x=-1 ?
The tangent of f(x)=x^5,\at x=-1 is y=5x+4