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

The tangent of f(x)=x^2,\at x=2x-1 is y=2(2x-1)x-12x^2+16x-5

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

tangent of

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

y = 2 (2 x - 1) x - 12 x^{2} + 16 x - 5

Solution steps

Find the tangent point:

(2 x - 1, (2 x - 1)^{2})

Find the slope of

f (x) = x^{2} : \frac{df ( x )}{d x} = 2 x

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 (2 x - 1)

Find the line with slope m=

2 (2 x - 1)

and passing through

(2 x - 1, (2 x - 1)^{2}) : y = 2 (2 x - 1) x - 12 x^{2} + 16 x - 5

y = 2 (2 x - 1) x - 12 x^{2} + 16 x - 5

\frac{d}{d y} (1 - y^{2})

n = 0 \sum \infty (- \frac{4}{9})^{n}

y = x^{\frac{1}{2}}, x = 0, x = 1

n = 1 \sum \infty \frac{5}{e ^{2 n}}

\frac{x y}{x + y}

What is the tangent of f(x)=x^2,\at x=2x-1 ?
The tangent of f(x)=x^2,\at x=2x-1 is y=2(2x-1)x-12x^2+16x-5