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

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

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

tangent of

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

y = - 2 x + 5

Solution steps

Find the tangent point:

(2, 1)

Find the slope of

f (x) = [(x - 1)^{2} - x]^{2} : \frac{df ( x )}{d x} = 2 (x^{2} - 3 x + 1) (2 x - 3)

Compute the slope of

f (x) = [(x - 1)^{2} - x]^{2}

x = 2 : m = - 2

Find the line with slope m=

- 2

and passing through

(2, 1) : y = - 2 x + 5

y = - 2 x + 5

\int \frac{x ^{2}}{225 - x ^{2}} d x

f (x) = cot^{2} (x) (csc (x))

f (x) = sin (cos (x))

\int 1 d x

\frac{\partial}{\partial x} (4 x^{2} y)

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