What is the tangent of f(x)=11+9x-x^3,\at x=1 ?

The tangent of f(x)=11+9x-x^3,\at x=1 is y=6x+13

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tangent of f(x)=11+9x-x^3,\at x=1

tangent of

f (x) = 11 + 9 x - x^{3}, at x = 1

y = 6 x + 13

Solution steps

Find the tangent point:

(1, 19)

Find the slope of

f (x) = 11 + 9 x - x^{3} : \frac{df ( x )}{d x} = - 3 x^{2} + 9

Compute the slope of

f (x) = 11 + 9 x - x^{3}

x = 1 : m = 6

Find the line with slope m=

6

and passing through

(1, 19) : y = 6 x + 13

y = 6 x + 13

\frac{\partial}{\partial x} (\frac{1}{( x - y ) ^{2}})

\frac{10 x ^{3} - 6 x ^{2} + 3}{4 x ^{2} - 6 x + 3}

x \to \infty lim (- \frac{1}{3} c e^{- 3 (x)})

\int (\frac{2}{x} - 4 e^{x}) d x

What is the tangent of f(x)=11+9x-x^3,\at x=1 ?
The tangent of f(x)=11+9x-x^3,\at x=1 is y=6x+13