What is the tangent of f(x)=sqrt(x^2-x+9),\at x=1 ?

The tangent of f(x)=sqrt(x^2-x+9),\at x=1 is y= 1/6 x+17/6

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tangent of f(x)=sqrt(x^2-x+9),\at x=1

tangent of

f (x) = x^{2} - x + 9, at x = 1

y = \frac{1}{6} x + \frac{17}{6}

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = x^{2} - x + 9 : \frac{df ( x )}{d x} = \frac{2 x - 1}{2 x ^{2} - x + 9}

Compute the slope of

f (x) = x^{2} - x + 9

x = 1 : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(1, 3) : y = \frac{1}{6} x + \frac{17}{6}

y = \frac{1}{6} x + \frac{17}{6}

\frac{10}{s ( s ^{2} + 8 s + 25 )}

\frac{\partial}{\partial x} (x arctan (yz))

x \to 0 lim (\frac{sin ( \frac{x}{2} )}{x})

x \to \infty lim ((1 + x)^{\frac{2}{x}})

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 2, y^{^{'}} (0) = 3

What is the tangent of f(x)=sqrt(x^2-x+9),\at x=1 ?
The tangent of f(x)=sqrt(x^2-x+9),\at x=1 is y= 1/6 x+17/6