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}

\frac{\partial ^{2}}{\partial x \partial y} (y^{5} sin (3 x))

n = 1 \sum \infty \frac{n 7 ^{n}}{n !}

\int 7 e^{5 x} d x

\int_{\frac{1}{2}}^{7} 12 x ln (2 x) d 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