What is the tangent of f(x)=sqrt(x^2-x+19),\at x=3 ?

The tangent of f(x)=sqrt(x^2-x+19),\at x=3 is y= 1/2 x+7/2

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tangent of f(x)=sqrt(x^2-x+19),\at x=3

tangent of

f (x) = x^{2} - x + 19, at x = 3

y = \frac{1}{2} x + \frac{7}{2}

Solution steps

Find the tangent point:

(3, 5)

Find the slope of

f (x) = x^{2} - x + 19 : \frac{df ( x )}{d x} = \frac{2 x - 1}{2 x ^{2} - x + 19}

Compute the slope of

f (x) = x^{2} - x + 19

x = 3 : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(3, 5) : y = \frac{1}{2} x + \frac{7}{2}

y = \frac{1}{2} x + \frac{7}{2}

\int \frac{1}{( x - 1 ) ^{\frac{1}{3}}} d x

f (x) = (x - 1) (x - 3)^{2}, at x = 4

(1 + 2 y) d x - (4 - x) d y = 0

f (x) = 13 x

\frac{d}{d x} (arcsin (7 x + 1))

What is the tangent of f(x)=sqrt(x^2-x+19),\at x=3 ?
The tangent of f(x)=sqrt(x^2-x+19),\at x=3 is y= 1/2 x+7/2