What is the tangent of f(x)=sqrt(x^2-x+10),\at x=3 ?

The tangent of f(x)=sqrt(x^2-x+10),\at x=3 is y= 5/8 x+17/8

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tangent of f(x)=sqrt(x^2-x+10),\at x=3

tangent of

f (x) = x^{2} - x + 10, at x = 3

y = \frac{5}{8} x + \frac{17}{8}

Solution steps

Find the tangent point:

(3, 4)

Find the slope of

f (x) = x^{2} - x + 10 : \frac{df ( x )}{d x} = \frac{2 x - 1}{2 x ^{2} - x + 10}

Compute the slope of

f (x) = x^{2} - x + 10

x = 3 : m = \frac{5}{8}

Find the line with slope m=

\frac{5}{8}

and passing through

(3, 4) : y = \frac{5}{8} x + \frac{17}{8}

y = \frac{5}{8} x + \frac{17}{8}

\int \frac{1}{v \cdot v ^{2} - 1} d v

f (x) = 2 x^{3}, (- 2, - 16)

\int cot (3 x) d x

ln (x x^{2} - 5)

\frac{7}{1 - x}

What is the tangent of f(x)=sqrt(x^2-x+10),\at x=3 ?
The tangent of f(x)=sqrt(x^2-x+10),\at x=3 is y= 5/8 x+17/8