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

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

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

tangent of

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

y = (2 a - 1) x - a^{2}

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, a^{2} - a)

Find the slope of

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

Compute the slope of

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

x = a : m = 2 a - 1

Find the line with slope m=

2 a - 1

and passing through

(a, a^{2} - a) : y = (2 a - 1) x - a^{2}

y = (2 a - 1) x - a^{2}

\int 6 x e^{3 x} d x

\frac{d}{d x} (x^{x^{2} + 1})

k = 0 \sum \infty \frac{1}{( k ! ) ^{2}}

y^{^{'}} - \frac{y}{t} = t e^{t}

\int - \frac{e ^{x} sin ( x )}{2} d x

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