What is the tangent of f(x)=9x^2-x^3,\at x=1 ?

The tangent of f(x)=9x^2-x^3,\at x=1 is y=15x-7

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tangent of f(x)=9x^2-x^3,\at x=1

tangent of

f (x) = 9 x^{2} - x^{3}, at x = 1

y = 15 x - 7

Solution steps

Find the tangent point:

(1, 8)

Find the slope of

f (x) = 9 x^{2} - x^{3} : \frac{df ( x )}{d x} = 18 x - 3 x^{2}

Compute the slope of

f (x) = 9 x^{2} - x^{3}

x = 1 : m = 15

Find the line with slope m=

15

and passing through

(1, 8) : y = 15 x - 7

y = 15 x - 7

\frac{\partial}{\partial y} (3 x + 2 y)

\frac{d y}{d x} = e^{3 x + y}

k = 1 \sum \infty 0. 5^{k} T

\int_{- 1}^{1} \frac{x}{5 - 4 x} d x

\int_{0}^{2} 2 - x^{2} d x

What is the tangent of f(x)=9x^2-x^3,\at x=1 ?
The tangent of f(x)=9x^2-x^3,\at x=1 is y=15x-7