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

The tangent of f(x)=x^2-x+1 is y=(2a_{0}-1)x-a_{0}^2+1

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

tangent of

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

y = (2 a_{0} - 1) x - a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - a_{0} + 1)

Find the slope of

f (x) = x^{2} - x + 1 : \frac{df ( x )}{d x} = 2 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 a_{0} - 1

Find the line with slope m=

2 a_{0} - 1

and passing through

(a_{0}, a_{0}^{2} - a_{0} + 1) : y = (2 a_{0} - 1) x - a_{0}^{2} + 1

y = (2 a_{0} - 1) x - a_{0}^{2} + 1

\frac{d}{d x} (\frac{2}{3} e^{x} + e^{- 2 x})

x \to 0 lim (1 - cos (x))

\frac{d}{d x} (\frac{x ^{- 2} + x ^{4} - 6}{3 x ^{3} + 4 x ^{4}})

e^{(4 x)}

\int x^{2} x^{3} + 3 d x

What is the tangent of f(x)=x^2-x+1 ?
The tangent of f(x)=x^2-x+1 is y=(2a_{0}-1)x-a_{0}^2+1