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

The tangent of f(x)=(5x-x^2)(5-x-x^2),\at x=1 is y=-3x+15

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

tangent of

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

y = - 3 x + 15

Solution steps

Find the tangent point:

(1, 12)

Find the slope of

f (x) = (5 x - x^{2}) (5 - x - x^{2}) : \frac{df ( x )}{d x} = 4 x^{3} - 12 x^{2} - 20 x + 25

Compute the slope of

f (x) = (5 x - x^{2}) (5 - x - x^{2})

x = 1 : m = - 3

Find the line with slope m=

- 3

and passing through

(1, 12) : y = - 3 x + 15

y = - 3 x + 15

f (x) = x - 3

\int_{- 4}^{2} f (x) + 2 x + 5 d x

\frac{1}{2} cos (2 t)

\int \frac{1}{u a ^{2} - u ^{2}} d u

\frac{\partial}{\partial z} (cos (x yz))

What is the tangent of f(x)=(5x-x^2)(5-x-x^2),\at x=1 ?
The tangent of f(x)=(5x-x^2)(5-x-x^2),\at x=1 is y=-3x+15