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

The tangent of f(x)=sqrt(x^2-x+2),\at x=2 is y= 3/4 x+1/2

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

tangent of

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

y = \frac{3}{4} x + \frac{1}{2}

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

f (x) = x^{2} - x + 2 : \frac{df ( x )}{d x} = \frac{2 x - 1}{2 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 = \frac{3}{4}

Find the line with slope m=

\frac{3}{4}

and passing through

(2, 2) : y = \frac{3}{4} x + \frac{1}{2}

y = \frac{3}{4} x + \frac{1}{2}

\int_{0}^{1} \frac{e ^{x}}{1 + e ^{2 x}} d x

x \to 3 lim (\frac{1}{3} x^{2} - 4 x + 10)

\frac{\partial}{\partial y} (\frac{3 y}{x})

\int \frac{x}{a + b x} d x

\int_{6}^{8} \frac{76}{( x - 6 ) ^{3}} d x

What is the tangent of f(x)=sqrt(x^2-x+2),\at x=2 ?
The tangent of f(x)=sqrt(x^2-x+2),\at x=2 is y= 3/4 x+1/2