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

The tangent of f(x)=[(x-2)^2-x]^2,\at x=3 is y=-4x+16

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tangent of f(x)=[(x-2)^2-x]^2,\at x=3

tangent of

f (x) = [(x - 2)^{2} - x]^{2}, at x = 3

y = - 4 x + 16

Solution steps

Find the tangent point:

(3, 4)

Find the slope of

f (x) = [(x - 2)^{2} - x]^{2} : \frac{df ( x )}{d x} = 2 (x^{2} - 5 x + 4) (2 x - 5)

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 4

Find the line with slope m=

- 4

and passing through

(3, 4) : y = - 4 x + 16

y = - 4 x + 16

4 - x^{2}, y = 0

\int_{3}^{6} 10000 e^{0.02 t} d t

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

x \to + 2 - lim (\frac{x}{x ^{2} - 4})

\int (1 + x^{2}) \cdot 1 + 64 x^{2} d x

What is the tangent of f(x)=[(x-2)^2-x]^2,\at x=3 ?
The tangent of f(x)=[(x-2)^2-x]^2,\at x=3 is y=-4x+16