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

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

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

tangent of

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

y = x + 1

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 1

Find the line with slope m=

1

and passing through

(1, 2) : y = x + 1

y = x + 1

\int_{- π}^{0} 0 d x

\int_{0}^{4} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x

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

f (x) = 3 x + \frac{5}{x}

x \to 3 + lim (x x - 3)

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