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

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

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

tangent of

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

y = 1

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

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

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 = 0

Find the line with slope m=

0

and passing through

(1, 1) : y = 1

y = 1

y^{^{'}} + \frac{1}{x} y = 4 x^{2}

\int - e^{4 x - 8} + 1 d x

\frac{d}{d x} (7 ln (2 x))

\int sin (4 x + π) d x

d er i v a t i v e \frac{1}{7 x}

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