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

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

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

tangent of

y = 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

y = x^{2} - x + 4 : \frac{d y}{d x} = \frac{2 x - 1}{2 x ^{2} - x + 4}

Compute the slope of

y = 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}

x \to - 0 lim (\frac{1}{1 + e ^{\frac{1}{x}}})

x \to - 6 lim (x - 10)

\int \frac{2 x ^{4}}{3 - 4 x ^{5}} d x

f (x) = x^{3} - 5 x^{2} - 2 x + 1, at x = - 2

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