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

The tangent of f(x)=sqrt(x^2-x+4),\at x=4 is y= 7/8 x+1/2

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tangent of f(x)=sqrt(x^2-x+4),\at x=4

tangent of

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

y = \frac{7}{8} x + \frac{1}{2}

Solution steps

Find the tangent point:

(4, 4)

Find the slope of

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

Compute the slope of

f (x) = x^{2} - x + 4

x = 4 : m = \frac{7}{8}

Find the line with slope m=

\frac{7}{8}

and passing through

(4, 4) : y = \frac{7}{8} x + \frac{1}{2}

y = \frac{7}{8} x + \frac{1}{2}

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What is the tangent of f(x)=sqrt(x^2-x+4),\at x=4 ?
The tangent of f(x)=sqrt(x^2-x+4),\at x=4 is y= 7/8 x+1/2