What is the tangent of f(x)=sqrt(4x^2-4x+1),\at x=1 ?

The tangent of f(x)=sqrt(4x^2-4x+1),\at x=1 is y=2x-1

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tangent of f(x)=sqrt(4x^2-4x+1),\at x=1

tangent of

f (x) = 4 x^{2} - 4 x + 1, at x = 1

y = 2 x - 1

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

f (x) = 4 x^{2} - 4 x + 1 : \frac{df ( x )}{d x} = \frac{2 ( 2 x - 1 )}{4 x ^{2} - 4 x + 1}

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 = 2

Find the line with slope m=

2

and passing through

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

y = 2 x - 1

\frac{\partial}{\partial x} (x sin (x + y) - cos (x + y))

y^{^{'}} = y^{4} \cdot (4 x^{2} + x)

(7, 4), (1, - 2)

\int (x^{2} - 2 x)^{2} d x

\int \frac{3}{( 4 + x ^{2} )} d x

What is the tangent of f(x)=sqrt(4x^2-4x+1),\at x=1 ?
The tangent of f(x)=sqrt(4x^2-4x+1),\at x=1 is y=2x-1